可能的重复项:
警告:mysql_fetch_array((:提供的参数不是有效的 MySQL 结果
在数据库中.php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;
}
}
?>
在索引中.php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
来自数据库.php。
我做错了什么?
尝试查找错误所在:
function getData() {
$data = array();
$sql = 'Select * From test';
$query = mysql_query($sql);
if(!$query)
{
echo 'Error: ' . mysql_error(); /* Check what is the error and print it */
exit;
}
while($row = mysql_fetch_array($query)) { /* Better use fetch array instead */
$data[] = array($row['id'], $row['name']);
}
return $data;
}
该错误告诉您由$query = mysql_query($sql);
执行的查询返回错误。它不返回零结果,而是返回一个错误,该错误表明名为 'databasename'
的数据库或该命名'test'
中的表不存在。
听起来查询
没有返回任何结果或常规查询错误,查询中的列和表是否存在,您与数据库的连接是否正常?