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Print filename with awk

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# 1  
Old 02-27-2012
Print filename with awk
I can got the filename with this script. it's only show "-" in result.

Code:
cut -d , -f7  CSV_d.*  | awk 'OFS=":"{print FILENAME,substr($1,1,8),substr($1,9,2),substr($1,11,2),substr($1,13,2)}' | sort |uniq

# 2  
Old 02-27-2012
Assuming your 7th field in the csv file does not contain any white space:


Code:
awk 'OFS=":"{print FILENAME,substr($7,1,8),substr($7,9,2),substr($7,11,2),substr($7,13,2)} FS=,  CSV_d.* | sort |uniq

(Not tested, since no sample files put. )

Guru.
This User Gave Thanks to guruprasadpr For This Post:
before4
# 3  
Old 02-27-2012
Quote:
Originally Posted by guruprasadpr
Assuming your 7th field in the csv file does not contain any white space:


Code:
awk 'OFS=":"{print FILENAME,substr($7,1,8),substr($7,9,2),substr($7,11,2),substr($7,13,2)} FS=,  CSV_d.* | sort |uniq

(Not tested, since no sample files put. )

Guru.

Thanks you , working welll
Code:
awk -F, 'OFS=":"{print FILENAME,substr($7,1,8),substr($7,9,2),substr($7,11,2),substr($7,13,2)}'   CSV.* | sort |uniq

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