An Error..
#1

Hi Guyz,I want if the player is registered and he join backs i want "Welcome back,[Player Name Here]"in that login dialog box,But getting three same warnings,"Warning 202:number of arguments does not match definition".Help me..
Код:
ShowPlayerDialog(playerid, DIALOG_LOGIN, DIALOG_STYLE_INPUT,""COL_WHITE"Login",""COL_WHITE"Welcome back,%s",""COL_WHITE"Type your password below to login.","Login","Quit",PlayerInfo[playerid][pName],playerid);
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#2

You need to format it.

PHP код:
new
    
str[128]
;
format(strsizeof(str), ""COL_WHITE" Welcome back %s, Type your password below to login."PlayerInfo[playerid][pName]);
ShowPlayerDialog(playeridDIALOG_IDDIALOG_STYLE"caption"str"Button 1""Button 2"); 
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#3

Quote:
Originally Posted by Arthur Kane
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You need to format it.

PHP код:
new
    
str[128]
;
format(strsizeof(str), ""COL_WHITE" Welcome back %s, Type your password below to login."PlayerInfo[playerid][pName]);
ShowPlayerDialog(playeridDIALOG_IDDIALOG_STYLE"caption"str"Button 1""Button 2"); 
Arthur it is not showing player name in the box,My Code:
Код:
format(str, sizeof(str), ""COL_WHITE" Welcome back %s, Type your password below to login.", PlayerInfo[playerid][pName]);
	ShowPlayerDialog(playerid, DIALOG_LOGIN, DIALOG_STYLE_INPUT, ""COL_WHITE"Login", str, "Login", "Quit");
screenshot:
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#4

Код:
new str[128], pname[MAX_PLAYER_NAME];
GetPlayerName(playerid,pname,sizeof(pname));
format(str, sizeof(str), ""COL_WHITE" Welcome back %s, Type your password below to login.", pname); 

ShowPlayerDialog(playerid, DIALOG_LOGIN, DIALOG_STYLE_INPUT,""COL_WHITE"Login",str,""COL_WHITE"Type your password below to login.","Login","Quit");
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#5

It's Fixed,Thank you Guyz
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