无语的出错,mysql_fetch_array()
代码如下，求帮忙看一下吧!
<?php<br>
class Conn{<br>
private $dbhost = ""; //服务器<br>
private $dbuser = ""; //mysql帐号<br>
private $dbpsw = ""; //mysql密码<br>
private $dbchar = ""; //数据库编码类型<br>
private $dbname = ""; //数据库名称<br>
private $tablepre = ""; //表前缀<br>
private $conn = NULL; //连接对象 <br>
<br>
/**<br>
* 数据库链接初始化<br>
*/<br>
public function __construct(){<br>
$this->getConn();<br>
}<br>
<br>
public function getConn(){<br>
$db = require_once 'config/config.php';<br>
$this->dbhost = $db['dbhost'];<br>
$this->dbuser = $db['dbuser'];<br>
$this->dbpsw = $db['dbpsw'];<br>
$this->dbchar = $db['dbchar'];<br>
$this->dbname = $db['dbname'];<br>
$this->tablepre = $db['tablepre'];<br>
$this->conn = mysql_connect($this->dbhost,$this->dbuser,$this->dbpsw) or die(mysql_error()."<br/>Mysql连接失败!");<br>
mysql_select_db($this->dbname,$this->conn) or die(mysql_error()."<br/>数据库访问出错");<br>
mysql_query("set names ".$this->dbchar,$this->conn);<br>
}<br>
<br>
/**<br>
* 执行sql<br>
*/<br>
public function query($sql){<br>
return mysql_query($sql,$this->conn) or die(mysql_error()."<br/>SQL执行出错：$sql");<br>
}<br>
<br>
/**<br>
* 返回多条记录<br>
*/<br>
public function getDataArrays($sql,$type = MYSQL_BOTH){<br>
$result = $this->query($sql);<br>
$refArr = array();<br>
<span style="color: #FF0000;">while ($row = mysql_fetch_array($result,$type)){</span><br>
$refArr[] = $row;<br>
}<br>
return $refArr;<br>
}<br>
<br>
<br>
/**<br>
* 关闭数据库链接<br>
*/<br>
public function closeConn(){<br>
mysql_close($this->conn);<br>
}<br>
}
调用如下：
$conn = new Conn();<br> $sql = "select * from qj_content";<br> $contentList = $conn->getDataArrays($sql,MYSQL_ASSOC);<br> $conn->closeConn();
执行结果如下：
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\AppServ\www\qjcentury\conn.php on line 45
第45行代码在上面用红色标注了
MySQL
PHP
数据库连接
查询
