Replace a string in shell script using a variable

Question

I am using the below code for replacing a string inside a shell script.

echo $LINE | sed -e 's/12345678/"$replace"/g'

but it's getting replaced with $replace instead of the value of that variable.

Could anybody tell what went wrong?

user_10828773 · Answer

Found a graceful solution.

echo ${LINE//12345678/$replace}

user_15048947 · Answer

Let me give you two examples.

#!/bin/bash
LINE="12345678HI"
replace="Hello"
echo $LINE | sed -e "s/12345678/$replace/g"

LINE="12345678HI"
str_to_replace="12345678"
replace_str="Hello"
result=${str//$str_to_replace/$replace_str}
echo $result

Hope you will find it helpful!

user_281196 · Answer

I had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:

replace="salt & pepper"
echo "pass the salt" | sed "s/salt/${replace/&/\&}/g"

user_2102275 · Answer

Use this instead

echo $LINE | sed -e 's/12345678/$replace/g'

this works for me just simply remove the quotes