How to echo string with variable substitution in bash?

Question

I have a string like ${REPOSITORY}/company/api:${API_VERSION}. $REPOSITORY and $API_VERSION are shell variables.

$ echo ${DATA_API_VERSION}
latest
$ echo ${REPOSITORY}
com.company.repo

I want to get the interpolated string that shows the values of these variables and assign it to another variable.

This is what I get:

$ echo "$image"
${REPOSITORY}/company/api:${API_VERSION}

I want this:

com.company.repo/company/api:latest

Check that you’re using double quotes: `image="${REPOSITORY}/company/api:${API_VERSION}"`. I reproduced your example and got the result you want. — oneastok, Sep 16 '19 at 20:52

score 1 · Answer 1 · edited Sep 16 '19 at 20:12

1

You could use sed to search and replace the two variables.

#!/bin/bash

DATA_API_VERSION="latest"
REPOSITORY="com.company.repo"

image='${REPOSITORY}/company/api:${DATA_API_VERSION}'
sed -e "
    s/\${REPOSITORY}/$REPOSITORY/g
    s/\${DATA_API_VERSION}/$DATA_API_VERSION/g
" <<< "$image"

edited Sep 16 '19 at 20:12

John Kugelman

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answered Sep 16 '19 at 19:43

Sanketh

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Note that this will have trouble if the variables contain slashes, newlines, or other special characters. – John Kugelman Sep 16 '19 at 20:13

How to echo string with variable substitution in bash?

1 Answers1