How can I execute `date` inside of a cron tab job?
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
add a comment |
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
add a comment |
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
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
cron quoting command-substitution
edited Jan 20 '12 at 23:00
Gilles
543k12811001617
543k12811001617
asked Jan 20 '12 at 17:12
cwdcwd
14.1k53116157
14.1k53116157
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
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
theSHELL
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.
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
add a comment |
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
add a comment |
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:
add a comment |
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
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
add a comment |
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
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
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
theSHELL
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.
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
add a comment |
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
theSHELL
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.
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
add a comment |
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
theSHELL
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.
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
theSHELL
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.
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
add a comment |
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
add a comment |
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
add a comment |
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
add a comment |
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
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
answered Jul 3 '15 at 14:41
Trevi AwaterTrevi Awater
17315
17315
add a comment |
add a comment |
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:
add a comment |
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:
add a comment |
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:
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:
edited Aug 10 '17 at 10:24
Stéphane Chazelas
310k57586945
310k57586945
answered Jan 4 '16 at 8:41
Gawi - KaiGawi - Kai
6114
6114
add a comment |
add a comment |
In cron, you can use this simple syntax:
*/15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1
Output date format will retrun like cron_20180123.log
– bala4rtraining
Jan 24 '18 at 13:51
(1) What are you saying that hasn’t already been said by the accepted answer? (2) Your answer is much more complicated than the question. For example, you added the-d
option, which is not used in the question (and you did not explain it). How do you justify calling this “simple syntax”?
– G-Man
Dec 23 '18 at 6:27
add a comment |
In cron, you can use this simple syntax:
*/15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1
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
add a comment |
In cron, you can use this simple syntax:
*/15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1
In cron, you can use this simple syntax:
*/15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +%Y%m%d).log 2>&1
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
add a comment |
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
add a comment |
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
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
add a comment |
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
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
add a comment |
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
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
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
add a comment |
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
add a comment |
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