manipulate first field and then print rest of the line for csv files using awk

Question

I have many csv file in which I need to manipulate the first field which contains the date and then print the rest of the line. The files have varying field lengths.

My sample line is

"11.07.2016 00:00:00",DON1SOE02,PAPN,PAPN,OPEN1000,918945,

I have tried the code below

awk -F"," '{print "\""substr($1,08,4)"-"substr($1,5,2)"-"substr($1,2,2)substr(‌$1,12,9)"\","$0""}' file.csv

The result of this is

"2016-07-11 00:00:00","11.07.2016 00:00:00",DON1SOE02,PAPN,PAPN,OPEN1000,918945,

Is there any way to avoid printing the unmodified date field, i.e. "11.07.2016 00:00:00"

solution below has a way to skip the first column, but I would like to modify the first column then print the modification and then skip printing unmodified first column

Using awk to print all columns from the nth to the last

score 3 · Accepted Answer · edited Jan 18 '17 at 12:04

3

You must give the new value to your the first field: awk '{$1="..."}1. Otherwise the whole line $0 will stay unchanged:

awk -F, -v OFS=, '{$1="\""substr($1,8,4)"-"substr($1,5,2)"-"substr($1,2,2)""substr($1,12,9)"\""}1' file.csv

edited Jan 18 '17 at 12:04

fedorqui 'SO stop harming'

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answered Jan 18 '17 at 11:50

Casimir et Hippolyte

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@fedorqui: it's indeed more clear like this. Thanks. – Casimir et Hippolyte Jan 18 '17 at 12:09
Thanks for the great solution. – Santosh Pillai Jan 18 '17 at 12:14

manipulate first field and then print rest of the line for csv files using awk

1 Answers1