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How to split file into multiple files using awk based on 1 field in the file?

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# 1  
Old 07-19-2012
How to split file into multiple files using awk based on 1 field in the file?
Good day all

I need some helps,

say that I have data like below, each field separated by a tab
Code:
DATE            NAME    ADDRESS
15/7/2012     LX         a.b.c
15/7/2012     LX1        a.b.c
16/7/2012     AB         a.b.c
16/7/2012     AB2        a.b.c
15/7/2012     LX2        a.b.c
15/7/2012     LX3         a.b.c
17/6/2012     CD         a.b.c

How can split this file into

Option 1: 3 daily files
Code:
15/7/2012
16/7/2012
17/6/2012

Option 2: 2 monthly files
Code:
june2012
july2012

Last edited by Scott; 07-19-2012 at 05:01 AM.. Reason: lease use code tags
# 2  
Old 07-19-2012
Hi alexyyw,

Welcome to the forum.

You could try these options.

For option1:
Code:
awk '{split($1,a,"/"); x="_"; print > a[1] x a[2] x a[3]}' file


For option2:
Code:
awk '{split($1,a,"/"); x="_"; print > a[2] x a[3]}' file


Please use code-tags next time.
# 3  
Old 07-19-2012
Hi alexyyw,

clx was faster, but here you have other way to have the result :

Option 1 :
Code:
awk '$1 ~ /.*\/.*\/..../ { split($1,a,"/"); print >a[1]"_"a[2]"_"a[3]; }' alexyyw.inputfile

Option 2 :
Code:
awk '$1 ~ /.*\/.*\/..../ { split($1,a,"/"); print >strftime("%B%Y",mktime(sprintf("%04d %02d %02d 00 00 00",a[3],a[2],a[1]))); }' alexyyw.inputfile

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