In my user's crontab, on a CentOS system, I have something like this:

58 12 * * * cd /home/joe/dev; echo `pwd` | logger

This outputs in /var/log/messages:

Feb 19 12:58:01 srv01 joe: /home/joe

Instead of outputting:

Feb 19 12:58:01 srv01 joe: /home/joe/dev

Why is the directory not kept when outputting pwd?

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    possibly because the path /home/joe/dev doesn't exist - so the first command doesn't do anything.
    – AD7six
    Feb 19, 2015 at 18:53
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    What happens if you do sh -c 'cd /home/joe/dev; echo `pwd` | logger' instead?
    – Zoredache
    Feb 19, 2015 at 19:01
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    @Daniel That's not for the user's crontab though, only the system-wide stuff. Feb 19, 2015 at 19:17
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    Guys, I feel bad, I managed to make it work however I don't know exactly how. But I think I might have made the mistake of not always exiting vim when I saved the changes to my crontab. I was used to edit the crontab manually and this time I was using crontab -e.
    – Martinos
    Feb 20, 2015 at 0:21
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    @Martinos that should be in the question (and emphasised) as it makes every answer except BeepBeep's "I tried that and it works fine" answer meaningless, since they are explaining a problem that doesn't exist.
    – AD7six
    Feb 20, 2015 at 8:16

3 Answers 3


The commands are being run independently, so it is not the same as expecting them to run in the same shell/environment. You would probably be better served by creating a shell script that performs the desired commands and then calling the script from the crontab.

Update: The above was a guess, and/or proposed possibilty (as a comment so tersely and aggressively pointed out). So I decided to test on CentOS 6.6 and... I was wrong... The following crontab entry worked properly:

* * * * * echo \`pwd` |logger; cd /home; echo `pwd` |logger;

So that should have worked for you... I still maintain that you would be best serverd by calling scripts not individual commands in your crontab... but that is a preference (so please be kind in the comments...).

  • That sounds like a guess, to fit "observed" behavior.
    – AD7six
    Feb 19, 2015 at 19:01
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    Lots of 'answers' are guesses here. That isn't necessarily a bad thing. An educated guess can be very helpful in leading people to solving their problems.
    – Zoredache
    Feb 19, 2015 at 19:04
  • @BeepBeep is right. If you use ; in cron, there are two independent commands running. If you want to concatenate them, use &&.
    – ThoriumBR
    Feb 19, 2015 at 19:24
  • The line that you wrote in you post works. However for some reason cd to my /home/joe/dev doesn't work. It seems to be a permission issue even the user running the cron is the same as the owner of the /home/joe/dev.
    – Martinos
    Feb 19, 2015 at 20:34
  • As the only sane answer (especially in light of the question being a misunderstanding) : +1. Please correct the escaped backtick, and preferably put the update first with emphasis - it doesn't matter to future readers you wrote something different originally.
    – AD7six
    Feb 20, 2015 at 8:22

The following should work:

58 12 * * * (cd /home/joe/dev; echo `pwd` | logger)

By wrapping the commands in parentheses, you should force them all to run in the same subshell.

  • Parentheses are not needed. All commands are already running as in the same subshell. Jun 28, 2020 at 15:22

As per this Unix&Linux SE answer, you can replace ; with && as different versions of cron across different Linux distributions handle this type of functionality in different ways:

cd /home/joe/dev && echo `pwd` | logger
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    "need"? The linked answer says nothing about ; not working. "My car doesn't start" "My blue car starts, you therefore need a blue car" is not a useful answer. (Note that the Asker comments that the question is effectively nonsense). -1.
    – AD7six
    Feb 20, 2015 at 8:19
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    @AD7six point taken, answer edited :) Feb 20, 2015 at 8:22
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    Thanks for taking my comment constructively, DV removed. different versions of cron across different Linux distributions... - I don't know of any version of cron, on any distro, that doesn't understand a semicolon, and of course ; and && are not logical equivalents.
    – AD7six
    Feb 20, 2015 at 8:27

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