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awk -remove pattern from file

 
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# 1  
Old 01-18-2016
awk -remove pattern from file
I have a file like this - I want to remove the 2015 (or any four digit #) from column $4 so I can get:
Code:
Nov 05 1997 /ifs/inventory2/

for example. Im not sure how. Should I use an if statement with awk?

Code:
Jan 16 2015 23:45 /ifs/sql_file
Jan 16 2015 23:45 /ifs/sql_file
Nov 05 2015 1997 /ifs/inventory2/
Nov 05 2015 1997 /ifs/inventory8/
Nov 05 2015 1997 /ifs/inventory7/
Nov 05 2015 1997 /ifs/inventory6/
Nov 05 2015 1997 /ifs/inventory5/

What I have got is this:
Code:
/tmp/filez1 |gawk '{if ($4 ~/1997/) }

but I don't know how to remove the 2015
from the file when 1997 is present.


I don't want anything removed from lines that do not have 1997
Moderator's Comments:
Mod Comment Using CODE tags isn't that hard. Please read the personal mail that received along with your infraction and review the tutorial that is included.
Last edited by Don Cragun; 01-18-2016 at 04:17 PM.. Reason: Change HTML tag to CODE tag to match closing CODE tag; add CODE and ICODE tags.
# 2  
Old 01-18-2016
Hello newbie2010,

Kindly use code tags as per forum rules for your commands/Inputs/codes which you are using into your posts, following may help you in same.
Code:
awk '{if($4 ~ /[0-9][0-9][0-9][0-9]/){$4="\b"}} 1'  Input_file

Output will be as follows.
Code:
Jan 16 2015 23:45 /ifs/sql_file
Jan 16 2015 23:45 /ifs/sql_file
Nov 05 2015 /ifs/inventory2/
Nov 05 2015 /ifs/inventory8/
Nov 05 2015 /ifs/inventory7/
Nov 05 2015 /ifs/inventory6/
Nov 05 2015 /ifs/inventory5/

Thanks,
R. Singh
# 3  
Old 01-18-2016
Try:
Code:
awk '$4~/^[0-9][0-9][0-9][0-9]$/{sub($4 FS,x)}1' file

or, if your awk supports this (gawk 3.x needs --re-interval):
Code:
awk '$4~/^[0-9]{4}$/{sub($4 FS,x)}1' file

or just:
Code:
awk '{sub(/^[0-9]{4} $/,x,$4)}1' file


--
Quote:
Originally Posted by RavinderSingh13
[..]
Code:
awk '{if($4 ~ /[0-9][0-9][0-9][0-9]/){$4="\b"}} 1'  Input_file

[..]
Try redirecting that to a file. You will notice that the backspace characters are still there, so I do not see how this can be a valid solution, other than that it looks OK on the terminal/screen...
Last edited by Scrutinizer; 01-18-2016 at 05:07 PM..
This User Gave Thanks to Scrutinizer For This Post:
RavinderSingh13
# 4  
Old 01-18-2016
Code:
perl -pae 's/ $F[3]// if $F[3] == "1997"' newbie2010.file

Code:
Jan 16 2015 23:45 /ifs/sql_file
Jan 16 2015 23:45 /ifs/sql_file
Nov 05 2015 /ifs/inventory2/
Nov 05 2015 /ifs/inventory8/
Nov 05 2015 /ifs/inventory7/
Nov 05 2015 /ifs/inventory6/
Nov 05 2015 /ifs/inventory5/

You may change the 1997 string to any other number you want.
# 5  
Old 01-19-2016
Not sure I understood the request entirely and correctly, but try
Code:
awk '$4 == "1997" {if ($3 == 2015) {$3=""; sub (FS FS, FS)}; print}' file
Nov 05 1997 /ifs/inventory2/
Nov 05 1997 /ifs/inventory8/
Nov 05 1997 /ifs/inventory7/
Nov 05 1997 /ifs/inventory6/
Nov 05 1997 /ifs/inventory5/

 
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