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Grep script name with path

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# 1  
Old 11-19-2014
Grep script name with path
would like to grep everything from first occurrence of / until >(not includes)

example :

0 5 * * * /usr/local/bin/shell/test.sh >/tmp/test.log
Output would be : /usr/local/bin/shell/test.sh
# 2  
Old 11-19-2014
grep cannot extract substrings from lines...it works on whole lines and prints out whole lines but other like sed and awk can do so...
# 3  
Old 11-19-2014
Code:
echo '0 5 * * * /usr/local/bin/shell/test.sh >/tmp/test.log' | sed 's/^.[^\/]*//;s/ .*//'

I guess you'll want
Code:
crontab -l | sed 's/^.[^\/]*//;s/ .*//'

---------- Post updated at 08:58 PM ---------- Previous update was at 08:55 PM ----------

If you have the sample lines in a file, then
Code:
sed 's/^.[^\/]*//;s/ .*//' file

# 4  
Old 11-19-2014
Thanks and it worked
# 5  
Old 11-19-2014
Moderator's Comments:
Mod Comment Having two threads discussing the same problem is confusing for those volunteers trying to help you solve your problem and wastes everyone's time. It looks like you got different solutions, but they all work.

The other thread is here.

This thread is closed.
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