I created a small bash script that takes a file using wget and then processes it using a php. The code goes like this:

wget -U mozilla -P /home/logfetcher/ http://fakesite.com/log.`date -d 'yesterday' +%Y-%m-%d`.csv
/usr/bin/php csv-editor.php /home/logfetcher/log.`date -d 'yesterday' +%Y-%m-%d`.csv /home/logfetcher/sorted/log.`date -d 'yesterday' +%Y-%m-%d`.csv 3 9 7 0 2 1 5 11 12 13 && rm /home/logfetcher/*.csv

I tested it and it works without any problems, though when I added it to cron like this:

0 6 * * * /home/logfetcher/fetchlogs.sh

It downloads the file but the php seems to don't work at all (nor the rm, which I guess indicates an error trying to run the php file). I've been trying to think about why this could be failing and tried a few things, but I don't seem to find a proper answer to fix it. Any help would be appreciated!


What is cron log saying? Or syslog? If there is error with first part of command then second command (after &&) won't be executed. is script csv-editor.php in your PATH? If it isn't than you should run it with absolute path /x/y/my.script.

I'm pretty sure that script isn't in your path, so "/usr/bin/php /x/y/script.php ...." should resolve problem.

But first read the logs...

  • I cannot believe I didn't think about that! Adding the full path to the script indeed fixed it. Thanks a lot Neven Aug 7 '15 at 9:14
  • 1
    Yeah the simplest thing are always hardest to spot. If this helped you, then please mark my post as an answer. Cheers
    – Neven
    Aug 7 '15 at 9:17

There are 2 solutions. Either you add #! /bin/bash as a first line in your script or you start your crontab file with the following 2 lines:


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