Bash shell scripting to update filenames in loop

Question

I have to create a shell script to execute the following.

for file in /dir
do
    somecommand $file /newdir/outputname.txt
done

My problem is for each loop I have to give new name for "outputname.txt' , like 'output1.txt' in loop1 , 'output2.txt' in loop2 and so on. How to do that ? Any help would be appreciated

use a variable and increment it. Something like let i=$i+1 — rpax
– rpax, Commented Jun 11, 2014 at 19:28

xiaodongjie · Accepted Answer · 2014-06-11 19:33:14Z

1

I can do this such as following.

i=1
for file in /dir
do
    somecommand $file /newdir/output${i}.txt
    i=$(($i + 1))
done

answered Jun 11, 2014 at 19:33

xiaodongjie

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Comments

konsolebox · Accepted Answer · 2014-06-11 19:53:49Z

1

This would not overwrite files that already exists in /newdir.

i=0
for file in /dir/*; do
    until [[ ! -e /newdir/output$((++i)).txt ]]; do
        continue
    done
    somecommand "$file" "/newdir/output${i}.txt"
done

* Always place arguments with variables around double-quotes to prevent word splitting. (Specially not necessary in [[ ]])

edited Jun 11, 2014 at 19:53

answered Jun 11, 2014 at 19:42

konsolebox

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2 Comments

quapka Over a year ago

The code highlighting appears to catch wrong language(looks like C or something). I'd suggest adding <pre><code class="bash> ... </code></pre> to your answer, or only <code class="bash"> ... </code>.

konsolebox Over a year ago

@quapka Thanks the fix is good enough. (Update: I instead used )

Pankrates · Accepted Answer · 2014-06-11 19:33:33Z

0

To put @rpax suggestion into a full answer:

i=0
for file in /dir
do
    i=$(($i+1))
    mv $file "/${newdir}/output_${i}.txt"
done

answered Jun 11, 2014 at 19:33

Pankrates

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