How can I execute `date` inside of a cron tab job? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Community Moderator Election ResultsAdding timestamp into log file via cronjob commandMplayer cronjob doesn't workUsage of '%' in the crontabHow can I pass a filename containing percent signs (%) as a parameter to a shell script in cron?Using date variable with wget --post-dataCrontab /bin/sh syntaxShell script not running in crontabWhat is causing my “Unexpected EOF Error while looking for …” error?Cron Job every 55 minutescron logging but not working on some commandsWhy can't cron job find basic Linux commands?Cron job does not fire up after a timezone changecron runs job at unexpected timeswhy daily cron isn't running on CentOS 6?Getting cron to include date in error logCron tab to run a java fileCron job isn't writing to log fileWhy does my cron job not work?Cron job not running on libreelecPython script runs as bash command on terminal but not as a cron job
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How can I execute `date` inside of a cron tab job?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Community Moderator Election ResultsAdding timestamp into log file via cronjob commandMplayer cronjob doesn't workUsage of '%' in the crontabHow can I pass a filename containing percent signs (%) as a parameter to a shell script in cron?Using date variable with wget --post-dataCrontab /bin/sh syntaxShell script not running in crontabWhat is causing my “Unexpected EOF Error while looking for …” error?Cron Job every 55 minutescron logging but not working on some commandsWhy can't cron job find basic Linux commands?Cron job does not fire up after a timezone changecron runs job at unexpected timeswhy daily cron isn't running on CentOS 6?Getting cron to include date in error logCron tab to run a java fileCron job isn't writing to log fileWhy does my cron job not work?Cron job not running on libreelecPython script runs as bash command on terminal but not as a cron job
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
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
547k13011131629
547k13011131629
asked Jan 20 '12 at 17:12
cwdcwd
14.2k54117158
14.2k54117158
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
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
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
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 Mar 20 at 21:25
Kusalananda♦
141k17263439
141k17263439
answered Jan 20 '12 at 17:31
Adam ZalcmanAdam Zalcman
2,73611513
2,73611513
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
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
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
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
314k57594952
314k57594952
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
1,181724
1,181724
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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