sed substitution with Bash variables

Question

I am trying to change the values in a text file using sed in a Bash script with the line,

sed 's/draw($prev_number;n_)/draw($number;n_)/g' file.txt > tmp 

This will be in a for loop. Why is it not working?

Answer 1

Variables inside ' don't get substituted in Bash. To get string substitution (or interpolation, if you're familiar with Perl) you would need to change it to use double quotes " instead of the single quotes:

# Enclose the entire expression in double quotes $ sed "s/draw($prev_number;n_)/draw($number;n_)/g" file.txt > tmp  # Or, concatenate strings with only variables inside double quotes # This would restrict expansion to the relevant portion # and prevent accidental expansion for !, backticks, etc. $ sed 's/draw('"$prev_number"';n_)/draw('"$number"';n_)/g' file.txt > tmp  # A variable cannot contain arbitrary characters # See link in the further reading section for details $ a='foo bar' $ echo 'baz' | sed 's/baz/'"$a"'/g' sed: -e expression #1, char 9: unterminated `s' command 

Further Reading:

• Difference between single and double quotes in Bash
• Is it possible to escape regex metacharacters reliably with sed
• Using different delimiters for sed substitute command
• Unless you need it in a different file you can use the -i flag to change the file in place

Answer 2

Variables within single quotes are not expanded, but within double quotes they are. Use double quotes in this case.

sed "s/draw($prev_number;n_)/draw($number;n_)/g" file.txt > tmp 

You could also make it work with eval, but don’t do that!!