我正在开发Android应用程序项目,在这个项目中,我需要连接我的Android应用程序与PHP和MySQL一切都很好,但我收到一个错误的用户注册.php文件错误像这样
警告:mysqli_close()期望参数1为我自己,给定null 在第35行的C:\ xampp \ htdocs \ panel \ UserRegistration.php
我尝试了很多东西,但错误仍然存在。
以下是我的尝试:
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
include 'DatabaseConfig.php';
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
$id_name = $_POST['id'];
$U_name = $_POST['username'];
$password = $_POST['password'];
$cnic = $_POST['cnic'];
$email = $_POST['email'];
$phone = $_POST['phone'];
$CheckSQL = "SELECT * FROM Users WHERE email='$email'";
$check = mysqli_fetch_array(mysqli_query($con,$CheckSQL));
if(isset($check)){
echo 'Email Already Exist';
}
else{
$Sql_Query = "INSERT INTO Users (id,username,email,password_cnic,phone) values ('$id','$U_name','$email','$password','$cnic','$phone')";
if(mysqli_query($con,$Sql_Query))
{
echo 'Registration Successfully';
}
}
?>
答案 0 :(得分:0)
https://secure.php.net/manual/fr/mysqli.close.php
你必须这样做:mysqli_close($con);
:)
答案 1 :(得分:0)
在mysqli_close()函数中传递Connection对象。
喜欢:
"<?=\yii\helpers\Url::to(['phone/index'])?>"