Variable in regex replace pattern

Question

Hi,
I need to use a variable in a regex replace pattern - how can I do it? Specifically, I need to pass arguments to a shell script that then uses that argument in a replace pattern:
#!/bin/bash
#$1 now holds the argument
sed 's/searchpattern/replace_pattern_with_variable$1/g' file1 > file2
when I run this, the replace pattern uses $1 as a literal string "$1" and not the variable value. How can I change this?
Thanks!
Ingo

Answer 1

Hi Ingo,
   As Vid points out, the issue is that single quotes protect strings from shell interpretation. You need to have the dollar sign, '$', visible to the shell or it won't read what follows as a variable name. Using double quotes works because the shell "reads through" those.
   However, complex regular expressions can contain lots of characters that the shell interprets. These can be quoted individually by backslashes but the use of backslashes in regular expressions is complex enough without the addition of shell interpretation. I find it easiest to keep the single quotes and only expose the part of the string that the shell needs to interpret.
   The shell doesn't have a special string concatenation character. All you have to do is to put the strings beside each other with nothing in between and the shell will concatenate them. Therefore it's possible to write your example as:
sed 's/searchpattern/replace_pattern_with_variable'${1}'/g' file1 > file2
That is, one closes the single quote right before the variable and then resumes it immediately afterward. The shell will put these quoted strings together with the contents of the variable just as it would with double quotes but you still enjoy the protection of single quotes around the rest of the string!
Gary
~~~~
   P.S. Perl's master plan (or what passes for one) is to take
   over the world like English did. Er, as English did...
      -- Larry Wall in <[email protected]>