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print column that match reg expr

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# 1  
Old 05-28-2009
print column that match reg expr
Hi all,

I want to cut a column which match the regular expression "beta", if I don't know the column number?

cat test

alpha;beta;gamma
11;22;33
44;55;66
77;88;99



should be

command ....

beta
22
55
88

I tried the following command, but unfortunately the output is only the regular expression not all the column rows!

awk -F";" 'match($0,"beta",a){print a[0]}' test

beta

Any ideas how can I solve this??
# 2  
Old 05-28-2009
use NR==1, then a for loop using NF as counter, loop through the fields one by one, checking for "beta". Get the value of the field number to a variable. After that print that value for every line.
# 3  
Old 05-28-2009
I will try to create the loop!

Thanks
# 4  
Old 05-29-2009
Code:
nawk -F";" '{
if(NR==1){
	for(i=1;i<=NF;i++){
		if($i=="beta"){
			n=i
			next
		}
	}
}
else{
	print $n
}
}' a.txt

# 5  
Old 05-29-2009
Thanks for you answer, but unfortunately I got the follow output:

Code:
~$ nawk -F";" '{if(NR==1){for(i=1;i<=NF;i++){if($i=="beta"){n=inext}}}else{print $n}}' test

11;22;33
44;55;66
77;88;99
# 6  
Old 05-29-2009
Code:
awk -F";" 'NR==1{for(i=1;i<=NF;i++)if ($i==...){n=i}}{print ....}' file

please familiarize yourself with the awk language by reading the gawk manual, available online.
# 7  
Old 05-29-2009
I solved this issue with the following script.

Code:
#!/bin/bash
#set -x

var=$(cat $2 | head -1 | sed 's/;/\n/g' | grep -n "$1" | cut -f1 -d":")

if [ -z $var ];then
echo "no string found"
else
cat $2 | cut -f$var -d";" 
fi


however ghostdog74 I going to take your advice about awk!

If you have any tutorials or tips regarding awk then please send it to me!

Thanks in advance!
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