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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result错误警告:mysql_fetch_assoc()在PHP
在db.php中我有:
<?php
class connect {
private $host = "localhost";
private $user = "root";
private $pass = "";
private $database = "databasename";
private $connect = null;
function connect() {
$this->connect = mysql_connect($this->host, $this->user, $this->pass) or die("Can't connect database");
mysql_select_db($this->database, $this->connect);
}
function getData() {
$data = array();
$sql = 'Select * From test';
$query = mysql_query($sql);
while($row = mysql_fetch_assoc($query)) {
$data[] = array($row['id'], $row['name']);
}
return $data;
}
}
?>
在index.php文件,我有:
<?php
include 'db.php';
$connect = new connect();
$connect->connect();
$data = $connect->getData();
$str = '';
foreach ($data as $dt) {
$str .= $dt[1];
}
echo $str;
?>
我收到以下错误: =>error: <b>Warning</b>: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource
from db.php。
我在做什么错?
我试图复制粘贴你的代码,没有发现错误!请检查您的数据库名称,数据库密码,数据库用户或表名。 – 2011-12-19 03:57:34
作为一个经验法则,你不应该命名与你的类相同的局部变量:'connect'。将'private $ connect'更改为(例如)'private $ connection'。然后你应该在你的查询中使用这个变量:$ query = mysql_query($ sql,$ this-> connection); – dar7yl 2011-12-19 11:40:58