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版主 sqlsrv 自个儿弄的简单类还需要您的帮助

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Release: 2016-06-13 12:14:18
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版主 sqlsrv 自己弄的简单类还需要您的帮助

<br />class DB_sqlsrv<br />{<br />    var $query;<br />	var $result;<br />	function DB_sqlsrv($text)<br />	{<br />		<br />				$serverName = "192.168.0.1";<br />				$connectionInfo = array(<br />				"UID"=>"sa",<br />				"PWD"=>"sa",<br />				"Database"=>"ttt"<br />				$conn = sqlsrv_connect( $serverName, $connectionInfo);<br />				$this->query=sqlsrv_query($conn, $text);<br />	}<br />	//以对象形式取得查询结果数据<br />	function Record()<br />	{<br />		$this->result=sqlsrv_fetch_object($this->query);<br />		return ($this->result)?($this->result):false;<br />	}<br />}<br />$sql=new DB_sqlsrv("select * from username");<br />$record=$sql->Record();<br />
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出错  PHP Warning:  sqlsrv_fetch_object(): 2 is not a valid ss_sqlsrv_stmt resource in E:\web\test\test\test.php on line 20
$this->result=sqlsrv_fetch_object($this->query);
这句有问题,我看看类好像没有问题的。请教谢谢。
------解决思路----------------------
$this->query=sqlsrv_query($conn, $text);
执行失败

你 var_dump($this->query); 看看是什么

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