我正在尝试使用PHP验证用户登录,但是我收到以下错误
警告:mysql_select_db()期望参数2是资源,对象 在第18行的C:\ xampp \ htdocs \ db \ sign_in_connect.php中给出
警告:mysql_fetch_array()期望参数1是资源, 第22行的C:\ xampp \ htdocs \ db \ sign_in_connect.php中给出的布尔值 抱歉,您的凭据无效,请重试。
这是我的PHP代码
<?php
try{
$db = mysqli_connect ('localhost', 'root', '', 'car_rental') or die ("SQL is Off");
}
catch (Exception $e){
echo "SQL is Off";
exit;
}
echo "success";
$email = $_POST["email"];
$pass = $_POST["pass"];
mysql_select_db("car_rental",$db);
$result = mysql_query("SELECT email, users FROM users WHERE email = $email");
$row = mysql_fetch_array($result);
if($row["email"]==$email && $row["pass"]==$pass)
echo"You are a validated user.";
else
echo"Sorry, your credentials are not valid, Please try again.";
?>
答案 0 :(得分:0)
用此替换您的代码。
AdView.setAdListener(new AdListener() {
// Implement AdListener
});