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/var bug #173136 21/03/07 03:39 AM
Joined: Oct 2005 Posts: 21 P poiuy_qwert OP Ameglian cow
OP poiuy_qwert Ameglian cow P Joined: Oct 2005 Posts: 21	If you set a variable with the /var command, then try to use that variable in another /var command, it will not work. For example: Code: alias badvar { var %a 1,%b $iif(%a,a,b) echo %b } Will give you * /echo: insufficient parameters. The only way to make it work is to do: Code: alias badvar { var %a 1,%b = $iif(%a,a,b) echo %b } - poiuy_qwert

Re: /var bug poiuy_qwert #173137 21/03/07 03:53 AM
Joined: Dec 2002 Posts: 2,962 Norwich, UK S starbucks_mafia Hoopy frood
starbucks_mafia Hoopy frood S Joined: Dec 2002 Posts: 2,962 Norwich, UK	Both of those work correctly for me. Neither is actually using the correct syntax for the var command though. It should be: Code: alias badvar { var %a = 1,%b = $iif(%a,a,b) echo %b } which works fine too. Spelling mistakes, grammatical errors, and stupid comments are intentional.

Re: /var bug poiuy_qwert #173152 21/03/07 01:49 PM
Joined: Oct 2004 Posts: 8,330 MA, USA Riamus2 Hoopy frood
Riamus2 Hoopy frood Joined: Oct 2004 Posts: 8,330 MA, USA	When using /var, the correct syntax is to always use an = sign. This prevents problems when you try setting a local variable (/var) with an identifier's value. Invision Support #Invision on irc.irchighway.net

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