I'm using sed to find and replace patterns within SAS files I have rather than changing them individually. The problem is I am trying to replace macro variables and when I use the ampersand it is not processing correctly.

Here's my code:
sed -ie 's/user=&uid./user=&sysuserid./g' *_table_*.sas

whenever I run this command it seems to append and do all kinds of funky stuff with the original text.

Question: How do I replace text that contains ampersands with sed command?

  • 7
    & is a special character so it needs to be escaped when used in the RHS - in your case you have to run 's/user=&uid./user=\&sysuserid./g' although I suspect you also need to escape the dot in the LHS to match a literal dot so you actually need 's/user=&uid\./user=\&sysuserid./g' Commented Jul 18, 2016 at 20:02
  • 1
    yes, '&' on the right basically means "everything that matched on the left", so escape it with '\'. You should make that the answer, rather than just a comment. Commented Jul 19, 2016 at 0:06
  • i didnt have to escape the first period...just a heads up in anyone else was curious :)
    – DukeLuke
    Commented Jul 20, 2016 at 0:50

1 Answer 1


& is special in the replacement text: it means “the whole part of the input that was matched by the pattern”, so what you're doing here replaces user=&uidX with user=user=&uidXsysuserid.. To insert an actual ampersand in the replacement text, use \&.

Another thing that looks wrong is that . in the search pattern stands for any character (except a newline), but the . at the end of the replacement text is a literal dot. If you want to replace only the literal string user=&uid., protect the . with a backslash.

sed -e 's/user=&uid\./user=\&sysuserid./g'

If you want to replace any one character and preserve it in the result, put the character in a group and use \1 in the replacement to refer to that group.

sed -e 's/user=&uid\(.\)/user=\&sysuserid\1/g'

In fact, given the repetition between the original text and the replacement, you should use groups anyway:

sed -e 's/\(user=&\)u\(id\.\)/\1sysuser\2/g'

i.e. “replace u by sysuser between user=& and id.”.

  • could you explain groups?
    – DukeLuke
    Commented Jul 20, 2016 at 0:53
  • 2
    @lucasdavis500 A group identifies a part of the pattern. It's delimited by backslash-parentheses. For example \(user=&\) is a pattern that matches user& and stores the match string as a group (group number 1 since it's the first group in the pattern). Then in the replacement \1 is replaced by the string stored for group number 1. Commented Jul 20, 2016 at 0:57
  • do you mean (user=&) stores user=&, not user&?
    – DukeLuke
    Commented Aug 17, 2016 at 12:40
  • @lucasdavis500 I don't understand your comment. The character = in regular expressions and in replacement text stands for itself, so user=& matches only user=&, and user=& in the replacement text yields user= followed by the part of the line matched by the regex. Commented Aug 17, 2016 at 13:27
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    @lucasdavis500 The = character does not have any special meaning. user=& in the replacement text produces user= followed by the original matched text. user=\& in the replacement text produces user=& Commented Aug 18, 2016 at 18:15

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