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I am writing a sed command that should uncomment an entry in crontab on sun Solaris 10.

I have tried 2 ways and they are working on Ubuntu but they didn't work on Sun Solaris 10; it returns sed: illegal option -- E crontab: can't open your crontab file.

crontab -l | sed -E '/#* *([^ ]+  *){5}[^ ]*run_all.sh/s/^#* *//' | crontab -

also :

crontab -l | sed '/#* *\([^ ][^ ]*  *\)\{5\}[^ ]*run_all.sh\.sh/s/^#* *//' | crontab -

shell on crontab looks like :

###15 00 * * * /bill/u01/WORK/ALARMS/run_all.sh > /bill/u01/WORK/ALARMS/`date +\%Y\%m\%d\%H\%M\%S`_RUN_ALL_PROCEDURE.log
  • Do you really need anything more complicated than '/run_all\.sh/ s/^##* *//'? in particular, I don't see any value in using capture groups within the address pattern – steeldriver Jul 27 '19 at 15:19
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You probably shouldn't overcomplicate the regex. To remove any possible hashtags at the beginning of lines containing the string run_all.sh, you could do:

crontab -l | sed 's/^#*\(.*run_all\.sh\)/\1/' | crontab -

Unfortunately, I don't have a Solaris system at hand to test it.

  • thank you dear you helped me alot, what if want to lead run_all.sh with ###shell what should i change? – mashro3ak Jul 27 '19 at 16:50
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    You mean the reverse operation? Use 's/.*run_all\.sh/###&/' instead. The & is a shorthand for the matched regex in the replacement. – Freddy Jul 27 '19 at 17:10

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