Unable to replace string using sed

Question

I have a bash script that loops over a set of files, and replaces a string, in each file, with a version number.

#!/bin/bash

ANGULAR_APP_VERSION=6.6.6
echo $ANGULAR_APP_VERSION

declare -a arr=(
"test1.txt" 
"test2.txt"
)

for i in "${arr[@]}"
do
  sed 's/"@@BUILD_VERSION@@"/"'$ANGULAR_APP_VERSION'"/g' ${arr[$i]}
done

Everytime I run the script, it generates the following error:

./test1.txt: syntax error: operand expected (error token is "./test1.txt")

I don't understand why it is wrong. The files exist.

see also stackoverflow.com/questions/7680504/… which is one of the faqs listed in stackoverflow.com/tags/sed/info — Sundeep
– Sundeep, Commented Apr 2, 2018 at 12:37

Gilles Quénot · Accepted Answer · 2018-04-02 12:29:32Z

3

#!/bin/bash

ANGULAR_APP_VERSION="6.6.6"
echo "$ANGULAR_APP_VERSION"

arr=(
    "test1.txt" 
    "test2.txt"
)

for i in "${arr[@]}"; do
     sed -i "s/@@BUILD_VERSION@@/$ANGULAR_APP_VERSION/g" "$i"
done

($i is each value of the array one at a time in the iteration)

or using array key :

for i in "${!arr[@]}"; do
     sed -i "s/@@BUILD_VERSION@@/$ANGULAR_APP_VERSION/g" "${arr[i]}"
done

Learn how to quote properly in shell, it's very important :

"Double quote" every literal that contains spaces/metacharacters and every expansion: "$var", "$(command "$var")", "${array[@]}", "a & b". Use 'single quotes' for code or literal $'s: 'Costs $5 US', ssh host 'echo "$HOSTNAME"'. See http://mywiki.wooledge.org/Quotes
http://mywiki.wooledge.org/Arguments
http://wiki.bash-hackers.org/syntax/words

edited Apr 2, 2018 at 12:29

answered Apr 2, 2018 at 12:19

Gilles Quénot

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4 Comments

Nicolas El Khoury Over a year ago

Using: sed "s/@@BUILD_VERSION@@/$ANGULAR_APP_VERSION/g" "$i" works, but it doesn't update the files, it just prints out the modified sring using: sed "s/@@BUILD_VERSION@@/$ANGULAR_APP_VERSION/g" "${arr[i]}" generates the following: test1.txt: syntax error: invalid arithmetic operator (error token is ".txt")

Gilles Quénot Over a year ago

For sure, you never asked to. Just add -i switch. Post edited accordingly

Nicolas El Khoury Over a year ago

Thank you! It worked. I will accept this answer in 5 minutes, when it allows me to.

Eric Renouf Over a year ago

It's probably worth noting that not every implementation of sed supports -i or supports it with no arguments

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