What exploit are these user agents trying to use?What web servers are affected by this user agent exploit?What is SPL exploit?What kind of security injection are these traces of, SQL, javascript, or otherwise?Is it illegal to use Fake User-agents?Server attack attempts, what are they trying to achieve?Can I exploit Windows kernel from user-mode application?HTTP attack taking down PHP-FPMSegmentation fault trying to exploit printf vulnerabilityWhat web servers are affected by this user agent exploit?Which exploit and which payload use?Help on what to do with these suspicious logs

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

Malformed Address '10.10.21.08/24', must be X.X.X.X/NN or

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Arrow those variables!

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What exploit are these user agents trying to use?


What web servers are affected by this user agent exploit?What is SPL exploit?What kind of security injection are these traces of, SQL, javascript, or otherwise?Is it illegal to use Fake User-agents?Server attack attempts, what are they trying to achieve?Can I exploit Windows kernel from user-mode application?HTTP attack taking down PHP-FPMSegmentation fault trying to exploit printf vulnerabilityWhat web servers are affected by this user agent exploit?Which exploit and which payload use?Help on what to do with these suspicious logs






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








48















I just looked at my user agent tracking page on my site (archived on Yandex) and I noticed these user agents. I believe they are an attempt to exploit my server (Nginx with PHP). The 1 in front of it is just how many times the user agent was seen in the Nginx log. These are also shortened user agents and not long ones like Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/73.0.3683.86 Safari/537.36. I no longer have access to the logs as I presume this occurred sometime in January or February (my oldest logs are in March and I created the site in January).



1 Mozilla/5.9}print(238947899389478923-34567343546345);improve this question



















  • 2





    Does the user agent start with ()? If yes, its probably the ShellShock exploit

    – Ferrybig
    2 days ago






  • 1





    @Ferrybig The shellshock exploit has a very particular syntax: ():;; is what triggers it.

    – Nzall
    2 days ago






  • 1





    A related question is security.stackexchange.com/questions/184115 .

    – JdeBP
    yesterday











  • Anecdotally, I appreciate that the numbers used are "pretty big." I used to get false-positive results from a vulnerability scanner that would add two 3-digit numbers in its math-problem tests. It would then "match" the sum in a substring of the Content-Length header.

    – Michael
    yesterday











  • In Plesk there used to be a vulnerability that allowed to execute php code that was within logs. This doesn't seem like it, but the vector of attack looks similar

    – eithed
    13 hours ago













shareprint(238947899389478923-34567343546345);improve this question
















I just looked at my user agent tracking page on my site (archived on Yandex) and I noticed these user agents. I believe they are an attempt to exploit my server (Nginx with PHP). The 1 in front of it is just how many times the user agent was seen in the Nginx log. These are also shortened user agents and not long ones like Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/73.0.3683.86 Safari/537.36. I no longer have access to the logs as I presume this occurred sometime in January or February (my oldest logs are in March and I created the site in January).



1 Mozilla/5.9print(238947899389478923-34567343546345);{
1 Mozilla/5.9$print(238947899389478923-34567343546345)
1 Mozilla/5.9x22$print(238947899389478923-34567343546345)x22
1 Mozilla/5.9x22];print(238947899389478923-34567343546345);//
1 Mozilla/5.9x22


What exploit was attempted and how can I test to ensure these exploits are not usable?







exploit webserver web nginx






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









forest

39.3k18127139




39.3k18127139










asked Apr 2 at 18:44









SenorContentoSenorContento

358138




358138







  • 2





    Does the user agent start with ()? If yes, its probably the ShellShock exploit

    – Ferrybig
    2 days ago






  • 1





    @Ferrybig The shellshock exploit has a very particular syntax: ():;; is what triggers it.

    – Nzall
    2 days ago






  • 1





    A related question is security.stackexchange.com/questions/184115 .

    – JdeBP
    yesterday











  • Anecdotally, I appreciate that the numbers used are "pretty big." I used to get false-positive results from a vulnerability scanner that would add two 3-digit numbers in its math-problem tests. It would then "match" the sum in a substring of the Content-Length header.

    – Michael
    yesterday











  • In Plesk there used to be a vulnerability that allowed to execute php code that was within logs. This doesn't seem like it, but the vector of attack looks similar

    – eithed
    13 hours ago












  • 2





    Does the user agent start with ()? If yes, its probably the ShellShock exploit

    – Ferrybig
    2 days ago






  • 1





    @Ferrybig The shellshock exploit has a very particular syntax: ():;; is what triggers it.

    – Nzall
    2 days ago






  • 1





    A related question is security.stackexchange.com/questions/184115 .

    – JdeBP
    yesterday











  • Anecdotally, I appreciate that the numbers used are "pretty big." I used to get false-positive results from a vulnerability scanner that would add two 3-digit numbers in its math-problem tests. It would then "match" the sum in a substring of the Content-Length header.

    – Michael
    yesterday











  • In Plesk there used to be a vulnerability that allowed to execute php code that was within logs. This doesn't seem like it, but the vector of attack looks similar

    – eithed
    13 hours ago







2




2





Does the user agent start with ()? If yes, its probably the ShellShock exploit

– Ferrybig
2 days ago





Does the user agent start with ()? If yes, its probably the ShellShock exploit

– Ferrybig
2 days ago




1




1





@Ferrybig The shellshock exploit has a very particular syntax: ():;; is what triggers it.

– Nzall
2 days ago





@Ferrybig The shellshock exploit has a very particular syntax: ():;; is what triggers it.

– Nzall
2 days ago




1




1





A related question is security.stackexchange.com/questions/184115 .

– JdeBP
yesterday





A related question is security.stackexchange.com/questions/184115 .

– JdeBP
yesterday













Anecdotally, I appreciate that the numbers used are "pretty big." I used to get false-positive results from a vulnerability scanner that would add two 3-digit numbers in its math-problem tests. It would then "match" the sum in a substring of the Content-Length header.

– Michael
yesterday





Anecdotally, I appreciate that the numbers used are "pretty big." I used to get false-positive results from a vulnerability scanner that would add two 3-digit numbers in its math-problem tests. It would then "match" the sum in a substring of the Content-Length header.

– Michael
yesterday













In Plesk there used to be a vulnerability that allowed to execute php code that was within logs. This doesn't seem like it, but the vector of attack looks similar

– eithed
13 hours ago





In Plesk there used to be a vulnerability that allowed to execute php code that was within logs. This doesn't seem like it, but the vector of attack looks similar

– eithed
13 hours ago










5 Answers
5






active

oldest

votes


















59














It looks to be trying to exploit some form of command injection. As DarkMatter mentioned in his answer, this was likely a broad attempt to find any vulnerable servers, rather than targeting you specifically. The payload itself just appears to just be testing to see if the server is vulnerable to command injection. It does not appear to have any additional purpose.



In order to test if you would be affected by these specific payloads, the easiest way would be to send them to your own server, and see how they respond. Note, that I only say this because the payloads themselves are benign; I do not recommend doing this with all payloads.



My bet is that your server is not vulnerable, because I would have expected to see follow up request to actually exploit your server.






share|improve this answer


















  • 5





    Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

    – Mic
    yesterday






  • 1





    That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

    – Ruslan
    15 hours ago



















22














It is probably nothing. It seems like the broad spam of a scanner looking across the web for any website that evaluates and returns that subtraction when it shouldn't. It is a pretty common thing to see.






share|improve this answer






























    21














    The use of actual function names (e.g. print) suggests they're looking for websites that are using eval in some way (note that this could be PHP's eval(string $code), JavaScript's eval(string), and other scripting languages' equivalents).



    I note that the executable code appears immediately after the first version parameter after Mozilla/. This means the authors of this attack believe that enough websites in the wild are actually using eval as a (horrible) way of parsing a two-component (major.minor) version number.



    So I imagine vulnerable websites were doing something like this (pseudo-code):



    var userAgent = request.headers["User-Agent"];

    var indexOfVersion = userAgent.indexof( '/' );
    var indexOfVersionEnd = userAgent.indexof( indexOfVersion , ' ' );

    var versionText = userAgent.substring( indexOfVersion + 1, indexOfVersionEnd );
    var versionNumber = eval( versionText ); // <------- this is the vulnerability!





    share|improve this answer
































      2














      it looks like they are trying to inject PHP code into log files. The idea being that if the sysadmin is using a PHP app to parse the logs, some might view the logfile as trusted (after all, the user does not normally get to directly alter the logfile) and therefore forego any sanitisation processes.



      If you are looking at your log files through a desktop or CLI text editor, you will never be vulnerable to this attack. If you use a PHP app, make sure it treats the logs as untrusted and sanitise it just like you would a normal user input field.






      share|improve this answer






























        1














        This is simple; they're trying PHP command injection. The process is to substitute a header (in this case the user agent field) with a mathematical expression, then to determine whether the code is being executed view the return value. If the code is executed, the return value will be the result of the expression, rather than the original expression. You'll notice the slightly spammy usage of open and close brackets, semicolons and other characters often used to fool interpreted languages into intepreting data as executable code. Nothing to worry about, automated vulnerability scans like this are par for the course nowadays.






        share|improve this answer








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        Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          59














          It looks to be trying to exploit some form of command injection. As DarkMatter mentioned in his answer, this was likely a broad attempt to find any vulnerable servers, rather than targeting you specifically. The payload itself just appears to just be testing to see if the server is vulnerable to command injection. It does not appear to have any additional purpose.



          In order to test if you would be affected by these specific payloads, the easiest way would be to send them to your own server, and see how they respond. Note, that I only say this because the payloads themselves are benign; I do not recommend doing this with all payloads.



          My bet is that your server is not vulnerable, because I would have expected to see follow up request to actually exploit your server.






          share|improve this answer


















          • 5





            Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

            – Mic
            yesterday






          • 1





            That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

            – Ruslan
            15 hours ago
















          59














          It looks to be trying to exploit some form of command injection. As DarkMatter mentioned in his answer, this was likely a broad attempt to find any vulnerable servers, rather than targeting you specifically. The payload itself just appears to just be testing to see if the server is vulnerable to command injection. It does not appear to have any additional purpose.



          In order to test if you would be affected by these specific payloads, the easiest way would be to send them to your own server, and see how they respond. Note, that I only say this because the payloads themselves are benign; I do not recommend doing this with all payloads.



          My bet is that your server is not vulnerable, because I would have expected to see follow up request to actually exploit your server.






          share|improve this answer


















          • 5





            Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

            – Mic
            yesterday






          • 1





            That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

            – Ruslan
            15 hours ago














          59












          59








          59







          It looks to be trying to exploit some form of command injection. As DarkMatter mentioned in his answer, this was likely a broad attempt to find any vulnerable servers, rather than targeting you specifically. The payload itself just appears to just be testing to see if the server is vulnerable to command injection. It does not appear to have any additional purpose.



          In order to test if you would be affected by these specific payloads, the easiest way would be to send them to your own server, and see how they respond. Note, that I only say this because the payloads themselves are benign; I do not recommend doing this with all payloads.



          My bet is that your server is not vulnerable, because I would have expected to see follow up request to actually exploit your server.






          share|improve this answer













          It looks to be trying to exploit some form of command injection. As DarkMatter mentioned in his answer, this was likely a broad attempt to find any vulnerable servers, rather than targeting you specifically. The payload itself just appears to just be testing to see if the server is vulnerable to command injection. It does not appear to have any additional purpose.



          In order to test if you would be affected by these specific payloads, the easiest way would be to send them to your own server, and see how they respond. Note, that I only say this because the payloads themselves are benign; I do not recommend doing this with all payloads.



          My bet is that your server is not vulnerable, because I would have expected to see follow up request to actually exploit your server.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 2 at 20:12









          user52472user52472

          2,872816




          2,872816







          • 5





            Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

            – Mic
            yesterday






          • 1





            That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

            – Ruslan
            15 hours ago













          • 5





            Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

            – Mic
            yesterday






          • 1





            That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

            – Ruslan
            15 hours ago








          5




          5





          Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

          – Mic
          yesterday





          Note that when re-testing a payload in this way, you don't check that you weren't vulnerable at the time it occurred (when maybe some updates were not yet made): just that you are not vulnerable anymore. Your server could still have been compromised - though I don't say it necessary is the case here.

          – Mic
          yesterday




          1




          1





          That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

          – Ruslan
          15 hours ago






          That there are these particular (apparently unsuccessful) attempts in the log doesn't mean there hasn't been a successful one, which didn't get logged. Notice how some of these potential commands do have $..., others don't, yet others have x22 which is quotation mark " etc. ­— the server may have been immune to some combinations of quoting/evaluating while vulnerable to others.

          – Ruslan
          15 hours ago














          22














          It is probably nothing. It seems like the broad spam of a scanner looking across the web for any website that evaluates and returns that subtraction when it shouldn't. It is a pretty common thing to see.






          share|improve this answer



























            22














            It is probably nothing. It seems like the broad spam of a scanner looking across the web for any website that evaluates and returns that subtraction when it shouldn't. It is a pretty common thing to see.






            share|improve this answer

























              22












              22








              22







              It is probably nothing. It seems like the broad spam of a scanner looking across the web for any website that evaluates and returns that subtraction when it shouldn't. It is a pretty common thing to see.






              share|improve this answer













              It is probably nothing. It seems like the broad spam of a scanner looking across the web for any website that evaluates and returns that subtraction when it shouldn't. It is a pretty common thing to see.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Apr 2 at 19:29









              DarkMatterDarkMatter

              2,3081121




              2,3081121





















                  21














                  The use of actual function names (e.g. print) suggests they're looking for websites that are using eval in some way (note that this could be PHP's eval(string $code), JavaScript's eval(string), and other scripting languages' equivalents).



                  I note that the executable code appears immediately after the first version parameter after Mozilla/. This means the authors of this attack believe that enough websites in the wild are actually using eval as a (horrible) way of parsing a two-component (major.minor) version number.



                  So I imagine vulnerable websites were doing something like this (pseudo-code):



                  var userAgent = request.headers["User-Agent"];

                  var indexOfVersion = userAgent.indexof( '/' );
                  var indexOfVersionEnd = userAgent.indexof( indexOfVersion , ' ' );

                  var versionText = userAgent.substring( indexOfVersion + 1, indexOfVersionEnd );
                  var versionNumber = eval( versionText ); // <------- this is the vulnerability!





                  share|improve this answer





























                    21














                    The use of actual function names (e.g. print) suggests they're looking for websites that are using eval in some way (note that this could be PHP's eval(string $code), JavaScript's eval(string), and other scripting languages' equivalents).



                    I note that the executable code appears immediately after the first version parameter after Mozilla/. This means the authors of this attack believe that enough websites in the wild are actually using eval as a (horrible) way of parsing a two-component (major.minor) version number.



                    So I imagine vulnerable websites were doing something like this (pseudo-code):



                    var userAgent = request.headers["User-Agent"];

                    var indexOfVersion = userAgent.indexof( '/' );
                    var indexOfVersionEnd = userAgent.indexof( indexOfVersion , ' ' );

                    var versionText = userAgent.substring( indexOfVersion + 1, indexOfVersionEnd );
                    var versionNumber = eval( versionText ); // <------- this is the vulnerability!





                    share|improve this answer



























                      21












                      21








                      21







                      The use of actual function names (e.g. print) suggests they're looking for websites that are using eval in some way (note that this could be PHP's eval(string $code), JavaScript's eval(string), and other scripting languages' equivalents).



                      I note that the executable code appears immediately after the first version parameter after Mozilla/. This means the authors of this attack believe that enough websites in the wild are actually using eval as a (horrible) way of parsing a two-component (major.minor) version number.



                      So I imagine vulnerable websites were doing something like this (pseudo-code):



                      var userAgent = request.headers["User-Agent"];

                      var indexOfVersion = userAgent.indexof( '/' );
                      var indexOfVersionEnd = userAgent.indexof( indexOfVersion , ' ' );

                      var versionText = userAgent.substring( indexOfVersion + 1, indexOfVersionEnd );
                      var versionNumber = eval( versionText ); // <------- this is the vulnerability!





                      share|improve this answer















                      The use of actual function names (e.g. print) suggests they're looking for websites that are using eval in some way (note that this could be PHP's eval(string $code), JavaScript's eval(string), and other scripting languages' equivalents).



                      I note that the executable code appears immediately after the first version parameter after Mozilla/. This means the authors of this attack believe that enough websites in the wild are actually using eval as a (horrible) way of parsing a two-component (major.minor) version number.



                      So I imagine vulnerable websites were doing something like this (pseudo-code):



                      var userAgent = request.headers["User-Agent"];

                      var indexOfVersion = userAgent.indexof( '/' );
                      var indexOfVersionEnd = userAgent.indexof( indexOfVersion , ' ' );

                      var versionText = userAgent.substring( indexOfVersion + 1, indexOfVersionEnd );
                      var versionNumber = eval( versionText ); // <------- this is the vulnerability!






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited yesterday

























                      answered yesterday









                      The DThe D

                      1,1261920




                      1,1261920





















                          2














                          it looks like they are trying to inject PHP code into log files. The idea being that if the sysadmin is using a PHP app to parse the logs, some might view the logfile as trusted (after all, the user does not normally get to directly alter the logfile) and therefore forego any sanitisation processes.



                          If you are looking at your log files through a desktop or CLI text editor, you will never be vulnerable to this attack. If you use a PHP app, make sure it treats the logs as untrusted and sanitise it just like you would a normal user input field.






                          share|improve this answer



























                            2














                            it looks like they are trying to inject PHP code into log files. The idea being that if the sysadmin is using a PHP app to parse the logs, some might view the logfile as trusted (after all, the user does not normally get to directly alter the logfile) and therefore forego any sanitisation processes.



                            If you are looking at your log files through a desktop or CLI text editor, you will never be vulnerable to this attack. If you use a PHP app, make sure it treats the logs as untrusted and sanitise it just like you would a normal user input field.






                            share|improve this answer

























                              2












                              2








                              2







                              it looks like they are trying to inject PHP code into log files. The idea being that if the sysadmin is using a PHP app to parse the logs, some might view the logfile as trusted (after all, the user does not normally get to directly alter the logfile) and therefore forego any sanitisation processes.



                              If you are looking at your log files through a desktop or CLI text editor, you will never be vulnerable to this attack. If you use a PHP app, make sure it treats the logs as untrusted and sanitise it just like you would a normal user input field.






                              share|improve this answer













                              it looks like they are trying to inject PHP code into log files. The idea being that if the sysadmin is using a PHP app to parse the logs, some might view the logfile as trusted (after all, the user does not normally get to directly alter the logfile) and therefore forego any sanitisation processes.



                              If you are looking at your log files through a desktop or CLI text editor, you will never be vulnerable to this attack. If you use a PHP app, make sure it treats the logs as untrusted and sanitise it just like you would a normal user input field.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered yesterday









                              520520

                              51524




                              51524





















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                                  This is simple; they're trying PHP command injection. The process is to substitute a header (in this case the user agent field) with a mathematical expression, then to determine whether the code is being executed view the return value. If the code is executed, the return value will be the result of the expression, rather than the original expression. You'll notice the slightly spammy usage of open and close brackets, semicolons and other characters often used to fool interpreted languages into intepreting data as executable code. Nothing to worry about, automated vulnerability scans like this are par for the course nowadays.






                                  share|improve this answer








                                  New contributor




                                  Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                    1














                                    This is simple; they're trying PHP command injection. The process is to substitute a header (in this case the user agent field) with a mathematical expression, then to determine whether the code is being executed view the return value. If the code is executed, the return value will be the result of the expression, rather than the original expression. You'll notice the slightly spammy usage of open and close brackets, semicolons and other characters often used to fool interpreted languages into intepreting data as executable code. Nothing to worry about, automated vulnerability scans like this are par for the course nowadays.






                                    share|improve this answer








                                    New contributor




                                    Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






















                                      1












                                      1








                                      1







                                      This is simple; they're trying PHP command injection. The process is to substitute a header (in this case the user agent field) with a mathematical expression, then to determine whether the code is being executed view the return value. If the code is executed, the return value will be the result of the expression, rather than the original expression. You'll notice the slightly spammy usage of open and close brackets, semicolons and other characters often used to fool interpreted languages into intepreting data as executable code. Nothing to worry about, automated vulnerability scans like this are par for the course nowadays.






                                      share|improve this answer








                                      New contributor




                                      Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.










                                      This is simple; they're trying PHP command injection. The process is to substitute a header (in this case the user agent field) with a mathematical expression, then to determine whether the code is being executed view the return value. If the code is executed, the return value will be the result of the expression, rather than the original expression. You'll notice the slightly spammy usage of open and close brackets, semicolons and other characters often used to fool interpreted languages into intepreting data as executable code. Nothing to worry about, automated vulnerability scans like this are par for the course nowadays.







                                      share|improve this answer








                                      New contributor




                                      Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      share|improve this answer



                                      share|improve this answer






                                      New contributor




                                      Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      answered yesterday









                                      Steve GazzoSteve Gazzo

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                                      111




                                      New contributor




                                      Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.





                                      New contributor





                                      Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      Steve Gazzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.



























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