Re: regex subexpressions broken

[Fran Litterio]

Ortwin wrote:

> [[ "a b c" =~ "a (.) c" ]]
> echo ${BASH_REMATCH[1]}

Recently, a change was made to cause quotes to suppress the special meaning
of regular expression characters in the right-hand side of =~.  This is how
to do the above now:

   [[ "a b c" =~ a\ (.)\ c ]]

I submitted a patch to the Bash man page describing these new quoting
semantics.  Just in case it was missed, I've included it again (below).
--
Francis Litterio


--- bash.1.orig    2006-10-03 08:54:26.000000000 -0400
+++ bash.1    2007-03-24 17:15:06.459699200 -0400
@@ -686,8 +686,19 @@
.if n .sp 1
An additional binary operator, \fB=~\fP, is available, with the same
precedence as \fB==\fP and \fB!=\fP.
-When it is used, the string to the right of the operator is considered
-an extended regular expression and matched accordingly (as in
\fIregex\fP(3)).
+When this operaor is used, the string to the right of the operator is
+considered an extended regular expression and matched accordingly (as
+in \fIregex\fP(3)).  If the regular expression string is quoted with
+either single or double quotes, the special meaning of the regular
+expression characters (such as '.', '*', '+', '[', ']', '(', ')',
+etc.) is suppressed.  Thus, if the regular expression string contains
+whitespace, it is best to escape the whitespace characters using '\\'
+instead of quoting the entire regular expression string, as follows:
+.if t .sp 0.5
+.if n .sp 1
+  \fB[[\fP "$DATA" \fB=~\fP foo\\ bar.*bletch \fB]]\fP
+.if t .sp 0.5
+.if n .sp 1
The return value is 0 if the string matches
the pattern, and 1 otherwise.
If the regular expression is syntactically incorrect, the conditional