Replace a string in shell script using a variable
UnixShellSedUnix Problem Overview
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?
Unix Solutions
Solution 1 - Unix
If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.
Try:
echo $LINE | sed -e "s/12345678/${replace}/g"
Transcript:
pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _
Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.
Solution 2 - Unix
you can use the shell (bash/ksh).
$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc
Solution 3 - Unix
Not specific to the question, but for folks who need the same kind of functionality expanded for clarity from previous answers:
# create some variables
str="someFileName.foo"
find=".foo"
replace=".bar"
# notice the the str isn't prefixed with $
# this is just how this feature works :/
result=${str//$find/$replace}
echo $result
# result is: someFileName.bar
str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo $result
# result is: someFileName.sally because ".foo" was not found
Solution 4 - Unix
Found a graceful solution.
echo ${LINE//12345678/$replace}
Solution 5 - Unix
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"
Solution 6 - Unix
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)
Solution 7 - Unix
Let me give you two examples.
- Using sed:
#!/bin/bash
LINE="12345678HI"
replace="Hello"
echo $LINE | sed -e "s/12345678/$replace/g"
- Without Using sed:
LINE="12345678HI"
str_to_replace="12345678"
replace_str="Hello"
result=${str//$str_to_replace/$replace_str}
echo $result
Hope you will find it helpful!
Solution 8 - Unix
To let your shell expand the variable, you need to use double-quotes like
sed -i "s#12345678#$replace#g" file.txt
This will break if $replace contain special sed characters (#, \). But you can preprocess $replace to quote them:
replace_quoted=$(printf '%s' "$replace" | sed 's/[#\]/\\\0/g')
sed -i "s#12345678#$replace_quoted#g" file.txt
Solution 9 - Unix
I had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:
replace="salt & pepper"
echo "pass the salt" | sed "s/salt/${replace/&/\&}/g"
Solution 10 - Unix
Use this instead
echo $LINE | sed -e 's/12345678/$replace/g'
this works for me just simply remove the quotes
Solution 11 - Unix
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"