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Forums Bug Reports $replace(%!a, a, t) doesn't work.

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$replace(%!a, a, t) doesn't work. #96294 31/08/04 07:23 AM
Joined: Sep 2003 Posts: 38 North America O Olathe OP Ameglian cow
OP Olathe Ameglian cow O Joined: Sep 2003 Posts: 38 North America	'echo 4 -a $replace(%!a, a, t)' doesn't work. It should echo '%t'.

Re: $replace(%!a, a, t) doesn't work. #96295 31/08/04 07:49 AM
Joined: Jan 2003 Posts: 3,012 New Hampshire, USA KingTomato Hoopy frood
KingTomato Hoopy frood Joined: Jan 2003 Posts: 3,012 New Hampshire, USA	%!a is being evaluated as a variable. Is this something that shouldn't happen? there is an array of different characters that a variable may be composed of, ! being one of them. -KingTomato

Re: $replace(%!a, a, t) doesn't work. #96296 31/08/04 07:52 AM
Joined: Apr 2004 Posts: 871 The Netherlands Sat Hoopy frood
Sat Hoopy frood Joined: Apr 2004 Posts: 871 The Netherlands	Unlike identifiers, you cannot escape variables like that - mIRC will try to fetch the variable named "%!a", which was probably empty in your case. Use "% $+ a" instead. Saturn, QuakeNet staff

Re: $replace(%!a, a, t) doesn't work. #96297 31/08/04 10:58 AM
Joined: Dec 2002 Posts: 788 Wales, United Kingdom. C Coolkill Hoopy frood
Coolkill Hoopy frood C Joined: Dec 2002 Posts: 788 Wales, United Kingdom.	$replace($+(%,a), a, t) Would be the solution, as mentioned. Eamonn.

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