Why percent signs (%) do not work in crontab?
I'm writing files out to a log ran by a bash script using cron. The call on cron looks like this:
*/25 * * * * bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
But when I check the crontab it records as
*/25 * * * * bash script.sh > "/var/log/$(date +).log"
And it never writes the log file. Is there something I need to change to get cron to write the date?
linux bash date cron crontab
add a comment |
I'm writing files out to a log ran by a bash script using cron. The call on cron looks like this:
*/25 * * * * bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
But when I check the crontab it records as
*/25 * * * * bash script.sh > "/var/log/$(date +).log"
And it never writes the log file. Is there something I need to change to get cron to write the date?
linux bash date cron crontab
add a comment |
I'm writing files out to a log ran by a bash script using cron. The call on cron looks like this:
*/25 * * * * bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
But when I check the crontab it records as
*/25 * * * * bash script.sh > "/var/log/$(date +).log"
And it never writes the log file. Is there something I need to change to get cron to write the date?
linux bash date cron crontab
I'm writing files out to a log ran by a bash script using cron. The call on cron looks like this:
*/25 * * * * bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
But when I check the crontab it records as
*/25 * * * * bash script.sh > "/var/log/$(date +).log"
And it never writes the log file. Is there something I need to change to get cron to write the date?
linux bash date cron crontab
linux bash date cron crontab
edited Apr 25 '16 at 11:39
fedorqui
167k53341381
167k53341381
asked Apr 26 '13 at 14:04
Devin DixonDevin Dixon
3,7141656115
3,7141656115
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add a comment |
1 Answer
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It is a matter of escaping variables:
* * * * * /usr/bin/touch /tmp/$(date +%Y:%m).log
# ^ ^
worked to me.
From man 5 crontab
:
Percent-signs (%) in the command, unless escaped with backslash (), will be changed into newline characters, and all data after the first % will be sent to the command as standard input.
So
*/25 * * * * /bin/bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
# ^ ^ ^ ^ ^
should work.
Note I used /bin/bash
instead of just bash
.
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is a matter of escaping variables:
* * * * * /usr/bin/touch /tmp/$(date +%Y:%m).log
# ^ ^
worked to me.
From man 5 crontab
:
Percent-signs (%) in the command, unless escaped with backslash (), will be changed into newline characters, and all data after the first % will be sent to the command as standard input.
So
*/25 * * * * /bin/bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
# ^ ^ ^ ^ ^
should work.
Note I used /bin/bash
instead of just bash
.
add a comment |
It is a matter of escaping variables:
* * * * * /usr/bin/touch /tmp/$(date +%Y:%m).log
# ^ ^
worked to me.
From man 5 crontab
:
Percent-signs (%) in the command, unless escaped with backslash (), will be changed into newline characters, and all data after the first % will be sent to the command as standard input.
So
*/25 * * * * /bin/bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
# ^ ^ ^ ^ ^
should work.
Note I used /bin/bash
instead of just bash
.
add a comment |
It is a matter of escaping variables:
* * * * * /usr/bin/touch /tmp/$(date +%Y:%m).log
# ^ ^
worked to me.
From man 5 crontab
:
Percent-signs (%) in the command, unless escaped with backslash (), will be changed into newline characters, and all data after the first % will be sent to the command as standard input.
So
*/25 * * * * /bin/bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
# ^ ^ ^ ^ ^
should work.
Note I used /bin/bash
instead of just bash
.
It is a matter of escaping variables:
* * * * * /usr/bin/touch /tmp/$(date +%Y:%m).log
# ^ ^
worked to me.
From man 5 crontab
:
Percent-signs (%) in the command, unless escaped with backslash (), will be changed into newline characters, and all data after the first % will be sent to the command as standard input.
So
*/25 * * * * /bin/bash script.sh > "/var/log/$(date +%Y-%m-%d_%H:%M).log"
# ^ ^ ^ ^ ^
should work.
Note I used /bin/bash
instead of just bash
.
edited Apr 25 '16 at 11:29
answered Apr 26 '13 at 14:21
fedorquifedorqui
167k53341381
167k53341381
add a comment |
add a comment |
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