$result = mysql_query("SELECT * FROM Cno");//查找 没有执行成功 mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\xampp\htdocs\4.php on line 12 请问是什么错误?如何改正?谢谢大家啊~ 代码如下: <?php $con = mysql_connect("localhost","root ",""); //通过服务器locahost建立连接,用户名为root,无密码 if (!$con) { die('Could not connect: ' . mysql_error()); }//如果不成功,显示错误 mysql_select_db("crms", $con);//选择数据库 $result = mysql_query("SELECT * FROM Cno");//查找 while($row = mysql_fetch_array($result))//打印 { echo $row['Cno'] . " " . $row['Cno']; echo "<br />"; } mysql_close($con); ?> |
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