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

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

Answer 2

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

Answer 3

Use this instead

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

this works for me just simply remove the quotes

Answer 4

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"

assuming you want the quotes put in. If you don't want the quotes, use:

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.