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wildcard in regex for awk

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# 1  
Old 09-21-2012
wildcard in regex for awk
Hello I have a file like :

Code:
20120918000001413 | 1.17.163.89 | iSelfcare | MSISDN | N 
20120918000001806 | 1.33.27.100 | iSelfcare | 5564 | N
....

I want to extract all lines that have on 4th field (considering "|" the separator ) something other than just digits. I want to do this using a awk one-liner ( I don;t want shell commands because combination of grep, while, cut.. etc takes way too much time)

I have this command :

Code:
awk -F'|' '$4 == " [A-Z]* "' target_file.txt

But it doesn't work.
Last edited by Scrutinizer; 09-21-2012 at 11:16 AM.. Reason: code tags also for data sample
# 2  
Old 09-21-2012
Try this one
Code:
awk -F\| '{if($4~/[a-zA-Z]+/){print}}' filename

This User Gave Thanks to senthilkumark For This Post:
black_fender
# 3  
Old 09-21-2012
Quote:
Originally Posted by senthilkumark
Try this one
Code:
awk -F\| '{if($4~/[a-zA-Z]+/){print}}' filename

Hello , thank you, exactly what I needed!
awk rules (but I'm too lazy to learn it :P )
# 4  
Old 09-21-2012
Code:
awk -F\| '$4!~/^[0-9 ]+$/' input_file


JUST ANYTHING other than numbers....including special characters like $%^ etc
This User Gave Thanks to msabhi For This Post:
black_fender
# 5  
Old 09-21-2012
Quote:
Originally Posted by msabhi
Code:
awk -F\| '$4!~/^[0-9 ]+$/' input_file

JUST ANYTHING other than numbers....including special characters like $%^ etc
Yes, this is more complete and accurate.
# 6  
Old 09-21-2012
Code:
awk -F\| '$4~/[^0-9 ]/' file

This User Gave Thanks to Scrutinizer For This Post:
black_fender
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