sed substitution with Bash variables
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?
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/'"$a"'/g'
sed: -e expression #1, char 9: unterminated `s' command
Further Reading:
- Difference between single and double quotes in Bash
- Is it possible to escape regex metacharacters reliably with sed
- Using different delimiters for sed substitute command
- Unless you need it in a different file you can use the -i flag to change the file in place
Variables within single quotes are not expanded, but within double quotes they are. Use double quotes in this case.
sed "s/draw($prev_number;n_)/draw($number;n_)/g" file.txt > tmp
You could also make it work with eval
, but don’t do that!!