Mysql querry problem - Printable Version
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Mysql querry problem -
smokks - 14.02.2016
Hi friends, Humm so i have a problem with the argument of a querry. I try to follow Mysql tutorial, but, when i just past what it's said, Pawno say me "argument type mismach 1"
Let's show the code
new string[256];
format(string,sizeof(string),"SELECT `Name` FROM `players` WHERE `Name` = '%s'",PlayerName);
mysql_query(string);
C:\**(62) : error 035: argument type mismatch (argument 1)
Pawn compiler 3.2.3664 Copyright © 1997-2006, ITB CompuPhase
1 Errors.
Thanks you and sorry for my english (i'm french :/ )
Re: Mysql querry problem -
K0P - 14.02.2016
Add the connection handle in mysql_query(); function
Код:
new mysql;
#define MYSQL_HOST ""
#define MYSQL_USER ""
#define MYSQL_DATABASE ""
#define MYSQL_PASSWORD ""
public OnGameModeInit()
{
mysql = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_DATABASE, MYSQL_PASSWORD);
return 1;
}
new string[256];
format(string,sizeof(string),"SELECT `Name` FROM `players` WHERE `Name` = '%s'",PlayerName);
mysql_query(/*here you need the connection handle*/ mysql ,string);
Re: Mysql querry problem -
smokks - 14.02.2016
Humm but what's the connection handle? Because i have hard about this but never get what is it
Re: Mysql querry problem -
AmigaBlizzard - 14.02.2016
It's the value returned by "mysql_connect" as shown in the example from "KOP".
PHP код:
mysql = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_DATABASE, MYSQL_PASSWORD);
"mysql" is your connection handle in this case.
You should include it to all your mysql commands to avoid confusion.