[boost] Re: [regex] Escaping a search string?

From: Edward Diener (eddielee_at_[hidden])
Date: 2003-08-07 14:37:56

Daniel Frey wrote:
> Edward Diener wrote:
>> You can turn on the literal flag type. All characters in your regular
>> expression are treated as literals.
>
> That doesn't help. Maybe an example clarifies what I need:
>
> std::string s = "1.30.0";
> boost::regex r( "^(.*)\s+(?:[Vv](?:ersion)?\s+" + s + ")\s*$" );
>
> I need a way to convert 's' to '1\.30\.0', not to escape the whole
> regex.

No such function exist in the regex++ library but it should be easy enough
to build one for yourself. In the regex++ doc, under the syntax topic there
is a list of all the characters that aren't literals. Put them in a static
string and write a transformation function which checks each character in
your source string and if it matches one in the static string, added a
backslash in front of it before putting it in a result string.

BTW, in your example above, you of course need two backslashes for every
backslash that occurs in your string literal.