Reputation: 4394
Putting $
inside square brackets doesn't work for grep.
~ $ echo -e "hello\nthere" > example.txt
~ $ grep "hello$" example.txt
hello
~ $ grep "hello[$]" example.txt
~ $
Is this a bug in grep or am I doing something wrong?
Upvotes: 14
Views: 5679
Reputation: 131
If you're trying to match end of line characters or the end of the string, you can use (|) like so "ABC($|\n)".
Upvotes: 10
Reputation: 1572
You can, however, use $
in a parenthesis grouping, which facilitates the use of |
(or), which can accomplish the same idea as a square bracket group.
Something like the following might be of interest to you:
~ $ cat example.txt
hello
there
helloa
hellob
helloc
~ $ grep "hello\($\|[ab]\)" example.txt
hello
helloa
hellob
Upvotes: 6
Reputation: 295579
That's what it's supposed to do.
[$]
...defines a character class that matches one character, $
.
Thus, this would match a line containing hello$
.
See the POSIX RE Bracket Expression definition for the formal specification requiring that this be so. Quoting from that full definition:
A bracket expression (an expression enclosed in square brackets, "[]" ) is an RE that shall match a single collating element contained in the non-empty set of collating elements represented by the bracket expression.
Thus, any bracket expression matches a single element.
Moreover, in the BRE Anchoring Expression definition:
- A dollar sign ( '$' ) shall be an anchor when used as the last character of an entire BRE. The implementation may treat a dollar sign as an anchor when used as the last character of a subexpression. The dollar sign shall anchor the expression (or optionally subexpression) to the end of the string being matched; the dollar sign can be said to match the end-of-string following the last character.
Thus -- as of BRE, the regexp format which grep
recognizes by default with no arguments -- if $
is not at the end of the expression, it is not required to be recognized as an anchor.
Upvotes: 11