PDO prepare()错误Php/Mysql

Jsh*_*hee 1 php mysql pdo

PHP

session_start(); 
$username = $_POST['regduser']; 
$userpass = md5($_POST['regdpass']); 
$sql = $sql->prepare("SELECT * from Students WHERE regduser='$username' and regdpass='$userpass'");
$sql->bindParam(':username', $username);
$sql->bindParam(':userpass', $userpass);
$stmnt->execute();
$result = mysql_query($sql); 
if (mysql_num_rows($result)!= 1) { 
 $error = "Login failed"; 
 #include "loginform.php"; 
} else { 
    echo "<h1>exists</h1>";
 #$_SESSION['regduser'] = "$username"; 
 #$_SESSION['ip'] = $_SERVER['REMOTE_ADDR']; 
 // any other data needed to navigate the site or 
 // to authenticate the user can be added here 
 #include "membersection.php"; 
}

?>
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HTML:

<form action="inc/check_regUsr.php" method="post" id="userLogon">
    <div class="field required">
        Username: <input type="text" name="regduser" tabindex="1" /><br />
        </div>
        <div class="field required">
        Password: <input type="password" name="regdpass" tabindex="2" /><br />
        </div>
        <input type="submit" name="submitUser" />
</form>
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致命错误:在第9 行的非对象上调用成员函数prepare()该行是:

$sql = $sql->prepare("SELECT * from Students WHERE regduser='$username' and regdpass='$userpass'");
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我在这做错了什么?!

Wri*_*ken 5

哦,从哪里开始......

  1. 数据库连接$sql在哪里?
  2. :placeholdername在预备陈述中使用,而不是$placeholdername.
  3. $sql如果您有数据库连接,则会覆盖,破坏数据库连接.
  4. $stmnt 不存在
  5. 这是什么mysql_query做的呢?你有3个选择:mysql,mysqli或PDO.坚持一个,不要混搭.