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
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
