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
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