Bash: iterate over several files and perform a different change for each file

Question

I want to change in several txt-files a variable. But this variable shouldn't be equal in every txt-file. I tried this, but it doesn't work.

for file in ./*.txt ;do
    for value in $(seq 1 5); do
        sed -i 's/x_max=.*/x_max='$value'/ ' $file
    done
done

So every x_max has got the value:5

Of course. You run each of the five replacements on each file. So they all end up with the last edit. — Jens
– Jens, Commented Mar 19, 2014 at 15:24
You are creating a loop before sed runs incrementing at every iteration. since the file is being over-written, every file will have x_max=5. — jaypal singh
– jaypal singh, Commented Mar 19, 2014 at 15:28

Jens · Accepted Answer · 2014-03-19 15:26:46Z

1

This should do the trick. Replace each file only once, with a different value each time.

value=1
for file in *.txt; do
  sed -i 's/x_max=.*/x_max='$value'/' $file
  value=$((value + 1))
done

answered Mar 19, 2014 at 15:26

Jens

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Comments

Saucier · Accepted Answer · 2014-03-19 15:28:23Z

0

That should do it - only iterate once and raise the counter by 1 after each run:

counter=1

for file in ./*.txt ;do
    sed -i 's/x_max=.*/x_max='$counter'/ ' $file
    (( counter++ ))
done

answered Mar 19, 2014 at 15:28

Saucier

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