如何在jqGrid中设置数据源 [英] how to set datasource in jqGrid
问题描述
我正在尝试使用jqGrid,但出现致命错误
i am trying to use jqGrid and i'm getting a fatal error
致命错误:消息无效"的未捕获异常"PDOException" 数据源名称"位于C:\ xampp \ htdocs \ psnm \ grid \ grid.php:8堆栈跟踪中:
Fatal error: Uncaught exception 'PDOException' with message 'invalid data source name' in C:\xampp\htdocs\psnm\grid\grid.php:8 Stack trace:
C:\ xampp \ htdocs \ psnm \ electionadmin.php(40): include('C:\ xampp \ htdocs ...')#2 {main}被抛出 C:\ xampp \ htdocs \ psnm \ grid \ grid.php在第8行
C:\xampp\htdocs\psnm\electionadmin.php(40): include('C:\xampp\htdocs...') #2 {main} thrown in C:\xampp\htdocs\psnm\grid\grid.php on line 8
,grid.php
的第8行是
$conn = new PDO("localhost",root,"");
我在localhost
上有服务器,我的数据库名称是psnm
,用户是root
,没有密码,有人可以帮我解决问题吗...
i have the server at localhost
my database name is psnm
, user is root
and no password, can anybody please help me find out the problem...
推荐答案
PDO的构造函数需要dsn作为第一个参数.请看下面的语法:
The constructor of the PDO needs the dsn as the first parameter. Look the syntax below:
PDO::__construct() ( string $dsn [, string $username [, string $password [, array $driver_options ]]] )
有关更多信息,请查阅PHP 手册.
Please consult PHP manual for more information.
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