Question

我这里有一个我正在处理的登录，我认为我的PHP存在一个小问题。

我的php连接到DB：

<?php
$db_name = "mydata";
$mysql_username = "root";
$mysql_password = "";
$server_name = "localhost";
$conn = mysqli_connect($server_name, $mysql_username, $mysql_password, $db_name);
if($conn){
    echo

 "Connection Succesful";
}
else{
    echo "Connection Not Succesful";
}
?>

我的php用于实际登录：

<?php
require "conn.php";
$Email = $_POST["emailPost"];
$Password = $_POST["passwordPost"];

$sql = "SELECT Password FROM users WHERE Email = '".$Email."' ";
$result = mysqli_query($conn, $sql);

if(mysqli_num_rows($result)>0){
    while($row = mysqli_fetch_assoc($result)){
        if($row == $Password){
            echo "login success";
        }
        else{
            echo "Password incorrect";
        }
    }
}else{
    echo "user not found";
}

?>

最后是我在C＃中使用的Coroutine

IEnumerator LoginAccount()
{
    WWWForm Form = new WWWForm();
    Form.AddField("emailPost", Email);
    Form.AddField("passwordPost", Password);

    WWW www = new WWW(LoginUrl, Form);
    yield return www;

    Debug.Log(www.text);
}

我每次都得到Password incorrect。我错过了什么？

Answer 1

找到它了！

var app = require('../app');
var debug = require('debug')('goWeb:server');
var http = require('http');

/**
 * Get port from environment and store in Express.
 */

var port = normalizePort(process.env.PORT || '3000');
app.set('port', port);

/**
 * Create HTTP server.
 */

var server = http.createServer(app);

/**
 * Listen on provided port, on all network interfaces.
 */

server.listen(port);
server.on('error', onError);
server.on('listening', onListening);

Answer 2

使用此代码（确保将此$row[Password]更新为与密码字段名称相同）

 <?php
    require "conn.php";
    $Email = $_POST["emailPost"];
    $Password = $_POST["passwordPost"];

    $sql = "SELECT Password FROM users WHERE Email = '".$Email."' ";
    $result = mysqli_query($conn, $sql);

    if(mysqli_num_rows($result)>0)
{
        while($row = mysqli_fetch_assoc($result))
{
            if($row['Password'] == $Password)
{
                echo "login success";
}
            else
{
                echo "Password incorrect";
            }}}
else
{
        echo "user not found";
    }

    ?>

PHP使用Unity和Mysql登录

2 个答案: