I want to print a line if i dont find the pattern using grep.! how is this possible?

ameylimaye · 09-28-2011, 10:34 AM

i want to print " -- " if the grep command is not able to find a pattern!

how can i do it?

so instead of disturbing my list of names...i can put "--" in that row.

Thank you in advance!

anomie · 09-28-2011, 10:44 AM

Hmm, grep(1). I think I'd use awk(1) or perl(1) instead for this one.

Code:

$ cat names.txt 
bill
barry
samantha
olga
sasha
nadya

Code:

$ awk '{ if($0 ~ /^sa/) print ; else print "--" }' names.txt 
--
--
samantha
--
sasha
--

ameylimaye · 09-28-2011, 11:00 AM

suppose i have a file in which there is a pattern

0001p0d5: vswr : 26.5[dB]

but sumtyms der is a pattern

0001p0d5: vswr : value not generated

so in the second case i need to print "--" instead of the value..

then?

anomie · 09-28-2011, 11:10 AM

That's a simple extension of my first example.

Code:

$ cat names.txt 
0001p0d5: vswr : 26.5[dB]
0001p0d5: vswr : 22.1[dB]
0001p0d5: vswr : value not generated
0001p0d5: vswr : 26.9[dB]

Code:

$ awk '{ if($0 ~ /value not generated/) print "--" ; else print }' names.txt
0001p0d5: vswr : 26.5[dB]
0001p0d5: vswr : 22.1[dB]
--
0001p0d5: vswr : 26.9[dB]

P.S. For the love of stout, frothy beer, please do not say things like "sumtyms der" here.

ameylimaye · 09-28-2011, 11:57 AM

ohh..! this will help me for sure.!

thanks a lot.!