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pass variable to awk

 
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# 1  
Old 12-18-2003
pass variable to awk
i would like to pass a variable to awk wherein the variable comes from external loop.

i tried this...

let x=0

until test $x -eq 32
do
cat file | awk '{ print $1 , "Number" , $($x) }' >> output
done

thanks,
# 2  
Old 12-19-2003
Try the following

cat file | awk -v xx=$x '{ print $1 , "Number" , xx }' >> output

- Finnbarr
# 3  
Old 12-19-2003
cat file | awk '{ print $1 , "Number" , $($x) }' >> output

-- Note: Awk is a little different from shell in that you do not need to use the $ sign to access a variable. The only time you would do so is if you wanted to get the value of a field variable. Example: $1 $2 ..$n where n is a value up to 199
# 4  
Old 02-11-2004
PHP sorry but not working
actually this is a sample of what i want to do:

from the sample code:

Code:
grep "keyword" filename | awk '{print $2}' >> outfile
grep "keyword" filename | awk '{print $3}' >> outfile
        .
        .
        .
grep "keyword" filename | awk '{print $32}' >> outfile

i want to shorten it to loop like this:
Code:
let x = 2

until test $x > 32
do

grep "keyword" filename | awk ...???

let x = x + 1

done

so you see i want to pass a variable to awk where the value will also be the variable for print.

thanks a lot
# 5  
Old 02-11-2004
Well, I'm assuming "keyword" is the same word in all those lines you have in your first section of code.... and I'm assuming the line it returns is gonna be a long line, and you're trying to print each part of the returned line on it's own line...

See if this works for you:
Code:
grep "keyword" filename | awk '{ for (i = 1; i <= NF; i++ ) print $i}' >> outfile

 
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