Question

我已经在托管网站上实现了数据库，而不是在使用MySQL的mac上。我使用数据库在IOS应用程序中进行SignUp / Login。数据库使用PHP代码进行连接。

这是PHP代码：

<?php

header('Content-type: application/json');
if($_POST) {
    $Nome   = $_POST['nome'];
    $Cognome   = $_POST['cognome'];
    $Email   = $_POST['email'];
    $DOB   = $_POST['dob'];
    $password   = $_POST['password'];
    $c_password = $_POST['c_password'];

    if($_POST['email']) {
        if ( $password == $c_password ) {

            $db_name     = '****************';
            $db_user     = '****************';
            $db_password = '*********************';
            $server_url  = 'localhost';

            $mysqli = new mysqli('localhost', $db_user, $db_name, $db_password);


            if (mysqli_connect_errno()) {
                error_log("Connect failed: " . mysqli_connect_error());
                echo '{"success":0,"error_message":"' . mysqli_connect_error() . '"}';
            } else {
                $stmt = $mysqli->prepare("INSERT INTO Users (nome, cognome, email, dob, password, c_password) VALUES (?, ?, ?, ?, ?, ?)");
                $password = md5($password);
                $stmt->bind_param('ssssss', $nome, $cognome, $email, $dob, $password, $c_password);


                $stmt->execute();

                if ($stmt->error) {error_log("Error: " . $stmt->error); }

                $success = $stmt->affected_rows;


                $stmt->close();


                $mysqli->close();
                error_log("Success: $success");

                if ($success > 0) {
                    error_log("User '$email' created.");
                    echo '{"success":1}';
                } else {
                    echo '{"success":0,"error_message":"Email Exist."}';
                }
            }
        } else {
            echo '{"success":0,"error_message":"Passwords does not match."}';
        }
    } else {
        echo '{"success":0,"error_message":"Invalid Email."}';
    }
}else {
    echo '{"success":0,"error_message":"Invalid Data."}';
}
?>

这是我在尝试数据库时看到的错误：

（HY000 / 2002）：无法通过socket＆＃39; /var/run/mysqld/mysqld.sock'连接到本地MySQL服务器; （2） /srv/disk8/2129310/www/haer2jkqlgjql.co.nf/signUp.php 20

此错误告诉我第20行有错误或有错误，但我不明白原因。

Answer 1

语法错误

$mysqli = new mysqli('localhost', $db_user, $db_name, $db_password);

正确的语法是

$mysqli = new mysqli('localhost',$db_user,$db_password, $db_name);

数据库连接出错：无法通过套接字

1 个答案: