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# 1  
Old 02-01-2016
Grep
When i do:
Code:
ls -ltr FILEPR*
700 FILEPR
0 FILPRHR
775 FILEPRDR

Code:
ls -tlr FILEPR* | grep -v 0

Result: 775 FILEPRD

I do not want 700 FILEPR to get removed; only 0 bit file.

I really appreciate any help you can provide.
Last edited by jim mcnamara; 02-01-2016 at 10:25 PM.. Reason: Fix code tags
# 2  
Old 02-01-2016
grep has the ability to search for characters at the beginning of a line. If you look at the examples section of the man page for grep you'll see it.

Code:
grep -v '^0'

should work.
This User Gave Thanks to [MA]Flying_Meat For This Post:
loktamann
# 3  
Old 02-01-2016
----
@loktamann: ls -ltr does not produce that kind of output. It is important that you state the data sample details accurately, so that forum members do not waste their valuable time.
----



If we take regular ls -ltr output as the data sample, then you could try this:
Code:
ls -tlr FILEPR* | awk '$5'

to obtain the non-empty file files

If you are only interested in the file names, then you could also try an all-shell approach:
Code:
for i in FILEPR*; do [ -s "$i" ] && echo "$i"; done

This User Gave Thanks to Scrutinizer For This Post:
loktamann
# 4  
Old 02-02-2016
I am an idiot. I forget about searching beginning or ending of word. grep -v '^0' works perfectly.

Scrutinizer, i love your code also.

Again, thank you so much for your time and assistance on this request.
Last edited by loktamann; 02-02-2016 at 06:26 AM.. Reason: spelling.
 
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