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llDialog

Smithy Zeno
ownes 9 pet cacti
Join date: 3 Aug 2006
Posts: 52
08-10-2007 11:10
How would I get this to work?

llDialog(llDetectedKey(i), (string)pm, MENU_MAIN, CHANNEL);

and in the listen statement I have

if (llGetSubString(llToLower(message),0,8) == "fg.setm. ";)
{
string pm = llGetSubString(message, 9, -1);
llSay(0, "Message set to "+(string)pm);
}

It just tells me in the error window 'llDialog: must supply a message'
I am probably doing something stupid, but if anyone could tell me how to make the dialog say the string that'd be great.
Pale Spectre
Registered User
Join date: 2 Sep 2005
Posts: 586
08-10-2007 11:53
From: http://www.lslwiki.net/lslwiki/wakka.php?wakka=llDialog

"Limits to message:

* message must be less than 512 characters and may not be empty, though it can contain a single space which will give the appearance of an empty message. If message is empty, the object will say "llDialog: must supply a message".
* If message is greater than 512 characters, the object will say "llDialog: message too long, must be less than 512 characters" and the dialog box will not appear.
* Newline "\n" and tab "\t" escape sequences are valid for formatting the message. Messages with more than 8 lines (including the result of text wrap) will cause a vertical scrollbar to be displayed inside the dialog. There isn't a way to change the actual size of the dialog box."


What does llOwnerSay((string)pm); report if you put it just prior to the llDialog. I'm guessing it shows nothing... an empty string.