How can I execute `date` inside of a cron tab job?












98















I want to create a log file for a cron script that has the current hour in the log file name. This is the command I tried to use:



0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


Unfortunately I get this message when that runs:



/bin/sh: -c: line 0: unexpected EOF while looking for matching ``'
/bin/sh: -c: line 1: syntax error: unexpected end of file


I have tried escaping the date part in various ways, but without much luck. Is it possible to make this happen in-line in a crontab file or do I need to create a shell script to do this?










share|improve this question





























    98















    I want to create a log file for a cron script that has the current hour in the log file name. This is the command I tried to use:



    0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


    Unfortunately I get this message when that runs:



    /bin/sh: -c: line 0: unexpected EOF while looking for matching ``'
    /bin/sh: -c: line 1: syntax error: unexpected end of file


    I have tried escaping the date part in various ways, but without much luck. Is it possible to make this happen in-line in a crontab file or do I need to create a shell script to do this?










    share|improve this question



























      98












      98








      98


      23






      I want to create a log file for a cron script that has the current hour in the log file name. This is the command I tried to use:



      0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


      Unfortunately I get this message when that runs:



      /bin/sh: -c: line 0: unexpected EOF while looking for matching ``'
      /bin/sh: -c: line 1: syntax error: unexpected end of file


      I have tried escaping the date part in various ways, but without much luck. Is it possible to make this happen in-line in a crontab file or do I need to create a shell script to do this?










      share|improve this question
















      I want to create a log file for a cron script that has the current hour in the log file name. This is the command I tried to use:



      0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


      Unfortunately I get this message when that runs:



      /bin/sh: -c: line 0: unexpected EOF while looking for matching ``'
      /bin/sh: -c: line 1: syntax error: unexpected end of file


      I have tried escaping the date part in various ways, but without much luck. Is it possible to make this happen in-line in a crontab file or do I need to create a shell script to do this?







      cron quoting command-substitution






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Jan 20 '12 at 23:00









      Gilles

      543k12811001617




      543k12811001617










      asked Jan 20 '12 at 17:12









      cwdcwd

      14.1k53116157




      14.1k53116157






















          5 Answers
          5






          active

          oldest

          votes


















          153














          Short answer:



          Escape the % as %:



          0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


          Long answer:



          The error message suggests that the shell which executes your command doesn't see the second back tick character:



          /bin/sh: -c: line 0: unexpected EOF while looking for matching '`'


          This is also confirmed by the second error message your received when you tried one of the other answers:



          /bin/sh: -c: line 0: unexpected EOF while looking for matching ')'


          The crontab manpage confirms that the command is read only up to the first unescaped % sign:




          The "sixth" field (the rest of the line) specifies the command to
          be run. The entire command portion of the line, up to a newline or
          % character, will be executed by /bin/sh or by the shell specified in
          the SHELL variable of the cronfile. Percent-signs (%) in the
          command, unless escaped with backslash (), will be changed into
          newline characters, and all data after the first % will be sent to
          the command as standard input.







          share|improve this answer


























          • Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

            – Tebe
            May 20 '15 at 6:24






          • 2





            @Копать_Шо_я_нашел cron will send an email with the error message,

            – Jasen
            Jan 1 '16 at 6:50






          • 3





            date +%Y %m %d %H:%M:%S-cronlog

            – DevilCode
            Apr 4 '16 at 13:36











          • This took me hours to find out... thanks

            – neurino
            Feb 19 at 9:28



















          7














          You can also put your commands into a shell file and then execute the shell file with cron.



          jobs.sh



          echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


          cron



          0 * * * * sh jobs.sh





          share|improve this answer































            6














            If you would like to make the date formatting string as a variable (to avoid duplicating the whole string), DO NOT escape % and DO NOT put it in $()



            For example, while declare the string, just write:



            DATEVAR=date +%Y%m%d_%H%M%S


            Then, write cron statement with $($VARIABLE_NAME) like this:



            * * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log


            Thanks to cyberx86, her/his answer at ServerFault might be more completed:






            share|improve this answer

































              2














              In cron, you can use this simple syntax:



              */15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1





              share|improve this answer


























              • Output date format will retrun like cron_20180123.log

                – bala4rtraining
                Jan 24 '18 at 13:51











              • (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                – G-Man
                Dec 23 '18 at 6:27



















              2














              All of the above answers use double quotes (not all of them worked for my setup). This worked for me:



              0 5 * * 3 /data/script.sh > /data/script_`date +%y%m%d`.log 2>&1





              share|improve this answer





















              • 1





                What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                – G-Man
                Dec 23 '18 at 6:27











              • The accepted answer simply doesn't work for me. This one does.

                – Manuel Schmitzberger
                Dec 23 '18 at 8:44











              Your Answer








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              5 Answers
              5






              active

              oldest

              votes








              5 Answers
              5






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              153














              Short answer:



              Escape the % as %:



              0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


              Long answer:



              The error message suggests that the shell which executes your command doesn't see the second back tick character:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching '`'


              This is also confirmed by the second error message your received when you tried one of the other answers:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching ')'


              The crontab manpage confirms that the command is read only up to the first unescaped % sign:




              The "sixth" field (the rest of the line) specifies the command to
              be run. The entire command portion of the line, up to a newline or
              % character, will be executed by /bin/sh or by the shell specified in
              the SHELL variable of the cronfile. Percent-signs (%) in the
              command, unless escaped with backslash (), will be changed into
              newline characters, and all data after the first % will be sent to
              the command as standard input.







              share|improve this answer


























              • Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

                – Tebe
                May 20 '15 at 6:24






              • 2





                @Копать_Шо_я_нашел cron will send an email with the error message,

                – Jasen
                Jan 1 '16 at 6:50






              • 3





                date +%Y %m %d %H:%M:%S-cronlog

                – DevilCode
                Apr 4 '16 at 13:36











              • This took me hours to find out... thanks

                – neurino
                Feb 19 at 9:28
















              153














              Short answer:



              Escape the % as %:



              0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


              Long answer:



              The error message suggests that the shell which executes your command doesn't see the second back tick character:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching '`'


              This is also confirmed by the second error message your received when you tried one of the other answers:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching ')'


              The crontab manpage confirms that the command is read only up to the first unescaped % sign:




              The "sixth" field (the rest of the line) specifies the command to
              be run. The entire command portion of the line, up to a newline or
              % character, will be executed by /bin/sh or by the shell specified in
              the SHELL variable of the cronfile. Percent-signs (%) in the
              command, unless escaped with backslash (), will be changed into
              newline characters, and all data after the first % will be sent to
              the command as standard input.







              share|improve this answer


























              • Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

                – Tebe
                May 20 '15 at 6:24






              • 2





                @Копать_Шо_я_нашел cron will send an email with the error message,

                – Jasen
                Jan 1 '16 at 6:50






              • 3





                date +%Y %m %d %H:%M:%S-cronlog

                – DevilCode
                Apr 4 '16 at 13:36











              • This took me hours to find out... thanks

                – neurino
                Feb 19 at 9:28














              153












              153








              153







              Short answer:



              Escape the % as %:



              0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


              Long answer:



              The error message suggests that the shell which executes your command doesn't see the second back tick character:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching '`'


              This is also confirmed by the second error message your received when you tried one of the other answers:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching ')'


              The crontab manpage confirms that the command is read only up to the first unescaped % sign:




              The "sixth" field (the rest of the line) specifies the command to
              be run. The entire command portion of the line, up to a newline or
              % character, will be executed by /bin/sh or by the shell specified in
              the SHELL variable of the cronfile. Percent-signs (%) in the
              command, unless escaped with backslash (), will be changed into
              newline characters, and all data after the first % will be sent to
              the command as standard input.







              share|improve this answer















              Short answer:



              Escape the % as %:



              0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


              Long answer:



              The error message suggests that the shell which executes your command doesn't see the second back tick character:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching '`'


              This is also confirmed by the second error message your received when you tried one of the other answers:



              /bin/sh: -c: line 0: unexpected EOF while looking for matching ')'


              The crontab manpage confirms that the command is read only up to the first unescaped % sign:




              The "sixth" field (the rest of the line) specifies the command to
              be run. The entire command portion of the line, up to a newline or
              % character, will be executed by /bin/sh or by the shell specified in
              the SHELL variable of the cronfile. Percent-signs (%) in the
              command, unless escaped with backslash (), will be changed into
              newline characters, and all data after the first % will be sent to
              the command as standard input.








              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 5 hours ago









              Kusalananda

              136k17257426




              136k17257426










              answered Jan 20 '12 at 17:31









              Adam ZalcmanAdam Zalcman

              2,69611513




              2,69611513













              • Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

                – Tebe
                May 20 '15 at 6:24






              • 2





                @Копать_Шо_я_нашел cron will send an email with the error message,

                – Jasen
                Jan 1 '16 at 6:50






              • 3





                date +%Y %m %d %H:%M:%S-cronlog

                – DevilCode
                Apr 4 '16 at 13:36











              • This took me hours to find out... thanks

                – neurino
                Feb 19 at 9:28



















              • Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

                – Tebe
                May 20 '15 at 6:24






              • 2





                @Копать_Шо_я_нашел cron will send an email with the error message,

                – Jasen
                Jan 1 '16 at 6:50






              • 3





                date +%Y %m %d %H:%M:%S-cronlog

                – DevilCode
                Apr 4 '16 at 13:36











              • This took me hours to find out... thanks

                – neurino
                Feb 19 at 9:28

















              Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

              – Tebe
              May 20 '15 at 6:24





              Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png

              – Tebe
              May 20 '15 at 6:24




              2




              2





              @Копать_Шо_я_нашел cron will send an email with the error message,

              – Jasen
              Jan 1 '16 at 6:50





              @Копать_Шо_я_нашел cron will send an email with the error message,

              – Jasen
              Jan 1 '16 at 6:50




              3




              3





              date +%Y %m %d %H:%M:%S-cronlog

              – DevilCode
              Apr 4 '16 at 13:36





              date +%Y %m %d %H:%M:%S-cronlog

              – DevilCode
              Apr 4 '16 at 13:36













              This took me hours to find out... thanks

              – neurino
              Feb 19 at 9:28





              This took me hours to find out... thanks

              – neurino
              Feb 19 at 9:28













              7














              You can also put your commands into a shell file and then execute the shell file with cron.



              jobs.sh



              echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


              cron



              0 * * * * sh jobs.sh





              share|improve this answer




























                7














                You can also put your commands into a shell file and then execute the shell file with cron.



                jobs.sh



                echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


                cron



                0 * * * * sh jobs.sh





                share|improve this answer


























                  7












                  7








                  7







                  You can also put your commands into a shell file and then execute the shell file with cron.



                  jobs.sh



                  echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


                  cron



                  0 * * * * sh jobs.sh





                  share|improve this answer













                  You can also put your commands into a shell file and then execute the shell file with cron.



                  jobs.sh



                  echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log


                  cron



                  0 * * * * sh jobs.sh






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jul 3 '15 at 14:41









                  Trevi AwaterTrevi Awater

                  17315




                  17315























                      6














                      If you would like to make the date formatting string as a variable (to avoid duplicating the whole string), DO NOT escape % and DO NOT put it in $()



                      For example, while declare the string, just write:



                      DATEVAR=date +%Y%m%d_%H%M%S


                      Then, write cron statement with $($VARIABLE_NAME) like this:



                      * * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log


                      Thanks to cyberx86, her/his answer at ServerFault might be more completed:






                      share|improve this answer






























                        6














                        If you would like to make the date formatting string as a variable (to avoid duplicating the whole string), DO NOT escape % and DO NOT put it in $()



                        For example, while declare the string, just write:



                        DATEVAR=date +%Y%m%d_%H%M%S


                        Then, write cron statement with $($VARIABLE_NAME) like this:



                        * * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log


                        Thanks to cyberx86, her/his answer at ServerFault might be more completed:






                        share|improve this answer




























                          6












                          6








                          6







                          If you would like to make the date formatting string as a variable (to avoid duplicating the whole string), DO NOT escape % and DO NOT put it in $()



                          For example, while declare the string, just write:



                          DATEVAR=date +%Y%m%d_%H%M%S


                          Then, write cron statement with $($VARIABLE_NAME) like this:



                          * * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log


                          Thanks to cyberx86, her/his answer at ServerFault might be more completed:






                          share|improve this answer















                          If you would like to make the date formatting string as a variable (to avoid duplicating the whole string), DO NOT escape % and DO NOT put it in $()



                          For example, while declare the string, just write:



                          DATEVAR=date +%Y%m%d_%H%M%S


                          Then, write cron statement with $($VARIABLE_NAME) like this:



                          * * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log


                          Thanks to cyberx86, her/his answer at ServerFault might be more completed:







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Aug 10 '17 at 10:24









                          Stéphane Chazelas

                          310k57586945




                          310k57586945










                          answered Jan 4 '16 at 8:41









                          Gawi - KaiGawi - Kai

                          6114




                          6114























                              2














                              In cron, you can use this simple syntax:



                              */15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1





                              share|improve this answer


























                              • Output date format will retrun like cron_20180123.log

                                – bala4rtraining
                                Jan 24 '18 at 13:51











                              • (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                                – G-Man
                                Dec 23 '18 at 6:27
















                              2














                              In cron, you can use this simple syntax:



                              */15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1





                              share|improve this answer


























                              • Output date format will retrun like cron_20180123.log

                                – bala4rtraining
                                Jan 24 '18 at 13:51











                              • (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                                – G-Man
                                Dec 23 '18 at 6:27














                              2












                              2








                              2







                              In cron, you can use this simple syntax:



                              */15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1





                              share|improve this answer















                              In cron, you can use this simple syntax:



                              */15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1






                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited Jan 24 '18 at 14:41









                              Kevin Lemaire

                              1,184724




                              1,184724










                              answered Jan 24 '18 at 13:50









                              bala4rtrainingbala4rtraining

                              312




                              312













                              • Output date format will retrun like cron_20180123.log

                                – bala4rtraining
                                Jan 24 '18 at 13:51











                              • (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                                – G-Man
                                Dec 23 '18 at 6:27



















                              • Output date format will retrun like cron_20180123.log

                                – bala4rtraining
                                Jan 24 '18 at 13:51











                              • (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                                – G-Man
                                Dec 23 '18 at 6:27

















                              Output date format will retrun like cron_20180123.log

                              – bala4rtraining
                              Jan 24 '18 at 13:51





                              Output date format will retrun like cron_20180123.log

                              – bala4rtraining
                              Jan 24 '18 at 13:51













                              (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                              – G-Man
                              Dec 23 '18 at 6:27





                              (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”?

                              – G-Man
                              Dec 23 '18 at 6:27











                              2














                              All of the above answers use double quotes (not all of them worked for my setup). This worked for me:



                              0 5 * * 3 /data/script.sh > /data/script_`date +%y%m%d`.log 2>&1





                              share|improve this answer





















                              • 1





                                What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                                – G-Man
                                Dec 23 '18 at 6:27











                              • The accepted answer simply doesn't work for me. This one does.

                                – Manuel Schmitzberger
                                Dec 23 '18 at 8:44
















                              2














                              All of the above answers use double quotes (not all of them worked for my setup). This worked for me:



                              0 5 * * 3 /data/script.sh > /data/script_`date +%y%m%d`.log 2>&1





                              share|improve this answer





















                              • 1





                                What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                                – G-Man
                                Dec 23 '18 at 6:27











                              • The accepted answer simply doesn't work for me. This one does.

                                – Manuel Schmitzberger
                                Dec 23 '18 at 8:44














                              2












                              2








                              2







                              All of the above answers use double quotes (not all of them worked for my setup). This worked for me:



                              0 5 * * 3 /data/script.sh > /data/script_`date +%y%m%d`.log 2>&1





                              share|improve this answer















                              All of the above answers use double quotes (not all of them worked for my setup). This worked for me:



                              0 5 * * 3 /data/script.sh > /data/script_`date +%y%m%d`.log 2>&1






                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited Sep 25 '18 at 9:36

























                              answered Sep 25 '18 at 8:46









                              Manuel SchmitzbergerManuel Schmitzberger

                              1213




                              1213








                              • 1





                                What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                                – G-Man
                                Dec 23 '18 at 6:27











                              • The accepted answer simply doesn't work for me. This one does.

                                – Manuel Schmitzberger
                                Dec 23 '18 at 8:44














                              • 1





                                What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                                – G-Man
                                Dec 23 '18 at 6:27











                              • The accepted answer simply doesn't work for me. This one does.

                                – Manuel Schmitzberger
                                Dec 23 '18 at 8:44








                              1




                              1





                              What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                              – G-Man
                              Dec 23 '18 at 6:27





                              What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.)

                              – G-Man
                              Dec 23 '18 at 6:27













                              The accepted answer simply doesn't work for me. This one does.

                              – Manuel Schmitzberger
                              Dec 23 '18 at 8:44





                              The accepted answer simply doesn't work for me. This one does.

                              – Manuel Schmitzberger
                              Dec 23 '18 at 8:44


















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