Bash script sed

Question

I am trying to use sed in bash script as follows:

#!/bin/bash
for i in `seq 1 10`;
do
    j=$(($i-1))
    OLD="-option_something something/string1_${j}.txt"
    NEW="-option_somehting something/string1_${i}.txt"
    sed -e "s/$OLD/$NEW/g" file_to_edit.txt
    # sed -e "s/$OLD/$NEW/g" file_to_edit.txt > file_to_edit.txt.tmp && mv file_to_edit.txt.tmp file_to_edit.txt 
done

But I keep getting following error:

sed: -e expression #1, char 71: unknown option tos'`

I tried the commented line as well, but it does not work too.

It works fine on command line. I do not know what is the problem in script.

Any suggestions? Thanks.

larsks · Accepted Answer · 2016-10-25 03:01:42Z

2

You have a / in the value of OLD and NEW, which is the same character you're using as the delimiter in your sed expression. So the final expression ends up looking like:

sed -e "s/-option_something something/string1_${j}.txt/-option_somehting something/string1_${i}.txt/g"

Do you see all the / in there? Consider instead:

sed -e "s|$OLD|$NEW|g" file_to_edit.txt

You can use any character as the delimiter for sed's s command.

answered Oct 25, 2016 at 3:01

larsks

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1 Comment

algoProg Over a year ago

Additionally I has to use -i instead of -e as my file was not getting edited and saved but just displayed in console and original file intact.

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