Mysql Error
#1

Hello friends i am trying to make my own game mode as im a begginer i came up with a problem that i cant solve any help is welcome

my errorlist:
Код:
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(77) : warning 204: symbol is assigned a value that is never used: "mysql"
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(120) : warning 202: number of arguments does not match definition
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(125) : warning 202: number of arguments does not match definition
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(349) : warning 219: local variable "pNamee" shadows a variable at a preceding level
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(351) : error 012: invalid function call, not a valid address
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(351) : warning 215: expression has no effect
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(351) : error 001: expected token: ";", but found ")"
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(351) : error 029: invalid expression, assumed zero
C:\Users\voxclient\Desktop\sampsvr\gamemodes\gdrp.pwn(351) : fatal error 107: too many error messages on one line
my line 351:
Код:
format(passquery,sizeof(passquery),"SELECT uspass FROM users WHERE usid = '%s';",pNamee(playerid));
the code block that line is in:
Код:
if(dialogid == 10 && response == 1)
	{
  		new passquery[80],pNamee[24];
  		GetPlayerName(playerid,pNamee,24);
  		format(passquery,sizeof(passquery),"SELECT uspass FROM users WHERE usid = '%s';",pNamee(playerid));
  		mysql_query(passquery);
  		mysql_store_result();
		if(mysql_fetch_string() == inputtext[])
		{
		    mysql_free_result();
		    return 1;
		}
		else if(mysql_fetch_string() != inputtext[])
		{
		    Kick(playerid);
		    return 1;
		}
        }
I cant see the problem here anybody can help?

PS: I am using strickenkid's mysql plugin and i have no intention to change it.
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#2

Hm.
PHP код:
format(passquery,sizeof(passquery),"SELECT uspass FROM users WHERE usid = '%s';",pNamee(playerid)); 
to
PHP код:
format(passquerysizeof(passquery),"SELECT uspass FROM users WHERE usid = '%s'",pNamee); 
And I don't know if it's for all MySQL version, mysql_query is defined like this:
Код:
mysql_query(conhandle, query[], bool:use_cache = true)
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#3

Thanks it fixed the problem and i dont know about
Код:
mysql_query(conhandle, query[], bool:use_cache = true)
but this works just fine with strickenkid's plugin
Код:
mysql_query(queryname);
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#4

Dont you think its about time you update your mysql plugin? that plugin you're using has kinda alot of problems (at least i had)
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