How to replace and remove few junk characters from a specific field?

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How to replace and remove few junk characters from a specific field?

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09-30-2014

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How to replace and remove few junk characters from a specific field?

I would like to remove all characters starting with "%" and ending with ")" in the 4th field - please help!!

Code:

1412007819.864  /device/services/heartbeatxx    204     0.547%!i(int=0)        0.434   0.112
1412007819.866  /device/services/heartbeatxx    204     0.547%!i(int=1)        0.423   0.123
1412007819.867  /device/services/heartbeatxx    204     0.547%!i(int=5)        0.450   0.096
1

note: field separator is "\t"

Last edited by vgersh99; 09-30-2014 at 04:55 PM.. Reason: code tags, please!

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Code:

awk -F'\t' 'match($4,"%.*[)]") {$4=substr($4,1,RSTART-1) substr($4,RSTART+RLENGTH+1)}1' OFS='\t' myFile

vgersh99

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Maybe,

Code:

awk -F"\t" 'sub(/%.*\)/, "", $4) 1'  OFS='\t' filename

Last edited by Aia; 09-30-2014 at 05:12 PM.. Reason: Correct the output field

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Sorry Aia and vgresh99, thanks for trying but both solution did not work - appreciate any other help.

snemuk14

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Quote:

Originally Posted by snemuk14

Sorry Aia and vgresh99, thanks for trying but both solution did not work - appreciate any other help.

Help us to help you. Could you explain how did it fail?

I'm guessing that if it did not work is because you are not using tabs as you assume, but rather spaces expanding to behave like tabs.

However, as I said before. Tell us how it did not work.

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On Solaris use nawk instead of awk......

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Ok. I stored the first line in a file named x to test and tried it on a linux machine as below:

Code:

$ awk -F"\t" 'sub(/%.*\)/, "", $4) 1'  OFS='\t' x

Output shows the same :

Code:

1412007819.864  /device/services/heartbeatxx    204     0.547%!i(int=0)        0.434   0.112

so, it did not eliminate the junk characters from "%" till ")"

Last edited by Don Cragun; 09-30-2014 at 09:04 PM.. Reason: Add CODE tags.

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