[SOLVED] Bash string with many variables of which i only want one of expressed

Issue

I have a variable called boiler,

boiler='#!/bin/bash
source ../../functions.sh
current="${si1}"
ready custom
title
breadcrumbs \""$current"\" \"Options\"
# END OF BOILER (DO NOT REMOVE ABOVE CODE OR MODIFY IT)
'

ISSUE and EXPECTED OUTPUT

The issue is that i want everything to be ignored withing this string (aka printed raw) except for the ${si1} variable. How could I concatenate the first part the variable and then the rest of the string, while keeping it minimal and saving it back into the boiler variable?

Solution

You can delimit the string around ${si1}.

boiler='#!/bin/bash
source ../../functions.sh
current='"${si1}"'
ready custom
title
breadcrumbs \""$current"\" \"Options\"
# END OF BOILER (DO NOT REMOVE ABOVE CODE OR MODIFY IT)
'

This is ordinary string concatenation. The strings delimited with ' will be literal, while the string delimited with " will have the variable expanded.

Difference between single and double quotes in Bash

Answered By - Barmar

Answer Checked By - Cary Denson (WPSolving Admin)