On Sun, 2003-06-15 at 09:43, Steven Broos wrote:
> Hai
>
> I'm trying to do a regular expression, which contains '
> As far as I know, I have to backslash the ' but that gives me an error.
>
> expr that returns the part of a string between (' '):
> $ expr "dag('hello')goodafternoon" : '.*(\(.*\)).*'
> 'hello'
> $ expr "dag('hello')goodafternoon" : '.*('\(.*\)').*'
> 0
> $ expr "dag('hello')goodafternoon" : '.*(\'\(.*\)\').*'
> bash: syntax error near unexpected token `)'
> $ expr "dag('hello')goodafternoon" : '.*(''\(.*\)'').*'
> 'hello'
> $ expr "dag('hello')goodafternoon" : '.*(.\(.*\).).*'
> hello
>
> Anyone has an idea how to contain a single quote in the expression ?
> I managed to work around this using just a point (last command above)
> but I really want to check for the quote
>
> tia
> Steven
Double quoting the regexp will do it:
expr "dag('hello')goodafternoon" : ".*('\(.*\)').*"
Adolfo
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