Ubuntu – How to use variables in a sed command

bashcommand linesed

I tried the following code to replace QQ with ZZ, but it doesn't do what I want:

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

However, this code does what I want:

sed -i 's/QQ/ZZ/g' $file

How do I use variables in sed?

Best Answer

  • 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/.


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

    s,bert,Say hurrah to &: \0/,

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

    Say hurrah to bert: \0/
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