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

Answer 1

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

you can still use single quotes, but you have to "open" them when you want the variable expanded at the right place. otherwise the string is taken "literally" (as @paxdiablo correctly stated, his answer is correct as well)

Answer 2

Found a graceful solution.

echo ${LINE//12345678/$replace}

Answer 3

I prefer to use double quotes , as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion .

so use double quotes instaed.

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

Answer 4

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"

Answer 5

you can use the shell (bash/ksh).

$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc

Answer 6

Single quotes are very strong. Once inside, there's nothing you can do to invoke variable substitution, until you leave. Use double quotes instead:

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