How do I use variables in a sed command?

Solution 1:

The shell is responsible for expanding variables. When you use single quotes for strings, its contents will be treated literally, so sed now tries to replace every occurrence of the literal $var1 by ZZ.

Using double quotes

Use double quotes to make the shell expand variables while preserving whitespace:

sed -i "s/$var1/ZZ/g" "$file"

When you require the quote character in the replacement string, you have to precede it with a backslash which will be interpreted by the shell. In the following example, the string quote me will be replaced by "quote me" (the character & is interpreted by sed):

sed -i "s/quote me/\"&\"/" "$file"

Using single quotes

If you've a lot shell meta-characters, consider using single quotes for the pattern, and double quotes for the variable:

sed -i 's,'"$pattern"',Say hurrah to &: \0/,' "$file"

Notice how I use s,pattern,replacement, instead of s/pattern/replacement/, I did it to avoid interference with the / in \0/.

Example

The shell then runs the above command sed with the next arguments (assuming pattern=bert and file=text.txt):

-i
s,bert,Say hurrah to &: \0/,
text.txt

If file.txt contains bert, the output will be:

Say hurrah to bert: \0/

Solution 2:

We can use variables in sed using double quotes:

sed -i "s/$var/r_str/g" file_name

If you have a slash / in the variable then use different separator, like below:

sed -i "s|$var|r_str|g" file_name

Solution 3:

To expand (pun intended) on @mani's answer,

• this solution will work for regular expressions fed to other commands, such as perl
• use double quotes around the regex (yes, even under Unixy systems) in order to expand the variable as expected
• | may appear in your variable's value as well, so don't be scared to try other delimiters