Shell Programming and Scripting

hi all,

how to read the column and print the values under that column ...??

file1 have something like this

cat file1
=======
column1, column2,date,column3,column4.....
1, 23 , 12/02/2008,......
2, 45, 14/05/2008,.....
3, 56, 16/03/2008,.....


cat file2
=======
column1,date,column3,column4.....
1,12/02/2008, 34,......
2,14/05/2008,56, .....
3,16/03/2008,78......

i need output like this

cat file1
=====
date
12/02/2008
14/05/2008
16/03/2008

cat file2
=====
date
12/02/2008
14/05/2008
16/03/2008


only date filed column values ...

pls help ...

thx in advance..

Code:

awk -F"," '{print $3}'file1

Similarly for file2...

awk -F"," '{print $2}'file2

The o/p will appear on the screen....redirect to a file as per ur requirment

thx for respond,
this i knew , but i need a script to handle that column date

date column maybe $1 , $2 ,$3, $4.....

thx,

Hope this helps

Code:

awk -F"," '/date/{}{print $(NF-1)}' testfile

Please ignore this post

Actually you are on the right track there.

Code:

awk -F, 'NR==1{for (i=0; i<=NF; i++) { if ($i == "date") break; } }
{ print $i }'

Meanwhile i was trying...heres one approach...
but i guess ...not a gd one

Code:

awk -F"|" '{ 
	    split ($0,a,",")
	    for (i=0 ;i<=NF;i++) 
		{
		if (a[i] == "date")
		val =i 
		}
	   } { print $val}'

Thanks 2 all , its working fine...