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# 1  
Old 10-21-2008
Question Another sed question
Hello, I am very new to shell scripting and have a directory path such as:
/usr/dev/blah/Arch/release/812-1235-P05/files/list and I want to output:

812-1235-P05

I think using sed with a regex like [0-9]-[0-9]-[0-9]?[a-z] would be the way to go, but I am having much trouble getting it to work. Any suggestions? Thank you.
# 2  
Old 10-21-2008
you need to add some quantifiers and limiters in there
Code:
sed 's/.*\([0-9]\{3\}-[0-9]\{4\}-[0-9A-Z]\{3\}\).*/\1/'

which will look for anything followed by \( exactly 3 digits followed by a - followed by exactly 4 digits, followed by a - followed by a combination of 3 letters and/or digits \) followed by anything and replace it all with the part above in \( .. \)
Last edited by wempy; 10-21-2008 at 11:38 AM.. Reason: for clarity
# 3  
Old 10-21-2008
Thank you so much for your reply. Looks like it should work, I am piping
/usr/dev/blah/Arch/release/812-1235-P05/files/list which is from a previous grep command into sed so like
Code:
grep command_to_get_path | sed 's/.*\([0-9]\{3\}-[0-9]\{4\}-[0-9A-Z]\{3\}\).*/\1/'

and it just displays the whole path again, it doesn't strip out everything around the 812-1235-P05. Am I doing something completely wrong? I must be.

Thanks again!
# 4  
Old 10-21-2008
That is strange, if I pipe that string directly into sed on my machine I get the required result. Can you run the grep without the sed and show us the exact output of the grep (including all spacing and 'wierd' chars).

my sed version:

chris@druid: ~ $ sed --version
GNU sed version 4.1.5
Last edited by wempy; 10-21-2008 at 11:57 AM.. Reason: added sed version
# 5  
Old 10-21-2008
I got it, my directory path also has a timestamp at the end of it and I didn't enclose it with quotes and it used some <>'s Newbie mistake I guess. Thank you much for all your help...certainly made my day easier.
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