bash for loop, use input filename in the output filename

Question

I want to use my input filename (baseline.YYYYMM.tar) in my output filename (baseline.YYYYMM.var1.tar). I can process the input files but don't know how to pass the output filename I need to my cdo application:

#!/bin/bash

prefix="basename"
fndate=$(ls | grep tar|cut -c 10-15)
var="var1"
extension=".tar"
outputfile=$prefix $fndate $var $extension
for f in $(find . -name "*.tar" -print) ; do
cdo selname,var1 $f $outputfile
done

thanks

Can you edit your question to make it clearer what the actual problem is and how we can help you? — Tom Fenech
– Tom Fenech, Commented Jan 25, 2015 at 21:30
Don't parse ls and definitely not randomly on the output of an entire directory. If you have a file you want the info for use stat on that file specifically. — Etan Reisner
– Etan Reisner, Commented Jan 25, 2015 at 21:43

Walter A · Accepted Answer · 2015-01-25 21:54:46Z

1

(I agree with remarks about parsing ls.) Did you forget dots?

 outputfile=$prefix $fndate $var $extension

should be

outputfile=${prefix}.${fndate}.${var}${extension}

answered Jan 25, 2015 at 21:54

Walter A

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1 Comment

chuckfinley Over a year ago

Thx Walter. Parsing was meant as a placeholder for a safer way to provide input.

Tom Fenech · Accepted Answer · 2015-01-25 21:49:26Z

0

I'm not 100% sure I get what you're trying to do but if I'm right, you should be able to use something like this to get your output file name from the input:

var=var1
for f in *.tar; do
    output=$(awk -v s="$var" 'BEGIN{FS=OFS="."}{print $1, $2, s, $3}' <<<"$f")
    # use "$output" however you want, e.g.
    echo "$output"
done

This uses awk to split up the input file name $f and insert the shell variable $var in the middle. <<<"$f" is bash syntax, equivalent to echo "$f" | at the start of the command.

answered Jan 25, 2015 at 21:49

Tom Fenech

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