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printing only columns containing a determined character with awk

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# 1  
Old 06-19-2012
Question printing only columns containing a determined character with awk
Hello

I'm trying to extract two columns from a database using awk, the thing is i have a variable number of columns in each row, and I just need to get out the two first ones ending in ","

This is an example:
Code:
ABE, ABE V149 MAZIE ARD CYN ACY, ACY, , , ,TEC, , 5000, , , 1, ZNY,ZDC
ABE, ABE FJC V149 LHY ALB, ALB, , , , TEC, , 8000, , ,1 ,ZNY, ZBW

As you can see the number of elements between the two one I need is not allways the same. I thought maybe with a for (for all the rows ) and an if (to selec for the column) I may be able to get it, but it's getting quite complicated for my level of awk .....I couldn't find any example for this kind of construction (or maybe I dind't understand it, hehe)
¿is there another better way to do it?
¿should I use awk for this?

thank you very much in any case Smilie
Last edited by Franklin52; 06-20-2012 at 03:36 AM.. Reason: Please use code tags for data and code samples
# 2  
Old 06-19-2012
Does this do it?

Code:
awk -F , ' { print $1 "," a[split($2, a, " ")]; } ' input-file >output-file

Output will be one pair per line, separated by a comma.
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