Php、MySql和Webform

我已经花了两天多的时间试图弄明白这一点，我正在学习youtube教程，为我的Android应用程序提供了一个基本的登录，但在我这么做之前，我想测试一下.php脚本

require("config.inc.php");

if (!empty($_POST)) {
    //gets user's info based off of a username.
    $query = "SELECT id, username, passwrd
            FROM application_users
            WHERE
                username = :username
        ";
    $query_params = array(
        ':username' => $_POST['username']
    );

    try {
        $stmt   = $db->prepare($query);
        $result = $stmt->execute($query_params);
    }
    catch (PDOException $ex) {
        // For testing, you could use a die and message.
        //die("Failed to run query: " . $ex->getMessage());

        //or just use this use this one to product JSON data:
        $response["success"] = 0;
        $response["message"] = "Database Error1. Please Try Again!";
        die(json_encode($response));

    }

    //This will be the variable to determine whether or not the user's information is correct.
    //we initialize it as false.
    $validated_info = false;

    //fetching all the rows from the query
    $row = $stmt->fetch();
    echo $row;
    if ($row) {
        //if we encrypted the password, we would unencrypt it here, but in our case we just
        //compare the two passwords
        if ($_POST['password'] === $row['password']) {
            $login_ok = true;
        }
    }

    // If the user logged in successfully, then we send them to the private members-only page
    // Otherwise, we display a login failed message and show the login form again
    if ($login_ok) {
        $response["success"] = 1;
        $response["message"] = "Login successful!";
        die(json_encode($response));
    } else {
        $response["success"] = 0;
        $response["message"] = "Invalid Credentials!";
        die(json_encode($response));
    }
} else {
?>
    <h1>Login</h1>
    <form action="login.php" method="post">
        Username:<br />
        <input type="text" name="username" placeholder="username" />
        <br /><br />
        Password:<br />
        <input type="password" name="password" placeholder="password" value="" />
        <br /><br />
        <input type="submit" value="Login" />
    </form>
    <a href="register.php">Register</a>
    </form>
<?php

我想当我按下登录按钮时应该会成功，但我得到的凭据无效，并且我知道用户名和密码是正确的

下面是我的login.php脚本

require("config.inc.php");

if (!empty($_POST)) {
    //gets user's info based off of a username.
    $query = "SELECT id, username, passwrd
            FROM application_users
            WHERE
                username = :username
        ";
    $query_params = array(
        ':username' => $_POST['username']
    );

    try {
        $stmt   = $db->prepare($query);
        $result = $stmt->execute($query_params);
    }
    catch (PDOException $ex) {
        // For testing, you could use a die and message.
        //die("Failed to run query: " . $ex->getMessage());

        //or just use this use this one to product JSON data:
        $response["success"] = 0;
        $response["message"] = "Database Error1. Please Try Again!";
        die(json_encode($response));

    }

    //This will be the variable to determine whether or not the user's information is correct.
    //we initialize it as false.
    $validated_info = false;

    //fetching all the rows from the query
    $row = $stmt->fetch();
    echo $row;
    if ($row) {
        //if we encrypted the password, we would unencrypt it here, but in our case we just
        //compare the two passwords
        if ($_POST['password'] === $row['password']) {
            $login_ok = true;
        }
    }

    // If the user logged in successfully, then we send them to the private members-only page
    // Otherwise, we display a login failed message and show the login form again
    if ($login_ok) {
        $response["success"] = 1;
        $response["message"] = "Login successful!";
        die(json_encode($response));
    } else {
        $response["success"] = 0;
        $response["message"] = "Invalid Credentials!";
        die(json_encode($response));
    }
} else {
?>
    <h1>Login</h1>
    <form action="login.php" method="post">
        Username:<br />
        <input type="text" name="username" placeholder="username" />
        <br /><br />
        Password:<br />
        <input type="password" name="password" placeholder="password" value="" />
        <br /><br />
        <input type="submit" value="Login" />
    </form>
    <a href="register.php">Register</a>
    </form>
<?php

require（“config.inc.php”）；
如果（！空（$\u POST））{
//根据用户名获取用户信息。
$query=“选择id、用户名、密码”
来自应用程序用户
哪里
用户名=：用户名
";
$query_params=数组(
':username'=>$\u POST['username']
);
试一试{
$stmt=$db->prepare（$query）；
$result=$stmt->execute（$query\u参数）；
}
捕获（PDO异常$ex）{
//对于测试，可以使用骰子和消息。
//die（“无法运行查询：”..ex->getMessage（））；
//或者只使用这个来生成JSON数据：
$response[“success”]=0；
$response[“message”]=“数据库错误1。请重试！”；
die（json_编码（$response））；
}
//这将是确定用户信息是否正确的变量。
//我们将其初始化为false。
$validated_info=false；
//正在从查询中获取所有行
$row=$stmt->fetch（）；
echo$行；
如果（$行）{
//如果我们加密了密码，我们会在这里取消加密，但在我们的情况下，我们只是
//比较两个密码
如果（$_POST['password']===$row['password']））{
$login\u ok=true；
}
}
//如果用户成功登录，则我们会将其发送到“仅限私人成员”页面
//否则，我们将显示登录失败消息并再次显示登录表单
如果（$login\u ok）{
$response[“success”]=1；
$response[“message”]=“登录成功！”；
die（json_编码（$response））；
}否则{
$response[“success”]=0；
$response[“message”]=“无效凭据！”；
die（json_编码（$response））；
}
}否则{
?>
登录
用户名：




密码：





因此，当加载脚本并从远程MYSQL服务器输入值时，消息会以无效凭据的形式返回。我只想确保登录成功，然后再转到android部分，这本身就是一个很大的任务。
我还没有机会用真实的数据库测试它，但这应该会起作用。您必须我必须在文件顶部添加
require（“config.inc.php”）；
，并且我添加了一个自定义数据库连接。我还使用PDO，因此查询看起来可能与您目前使用的不同

<?php
// Database connection
try
{
    $db = new PDO('mysql:host=localhost;dbname=mydatabase', 'myusername', 'mypassword');
    $db->exec('SET CHARACTER SET UTF8');
}
catch (Exception $e)
{
    //Message in case of error when connecting to the database
    die('Erreur : ' . $e->getMessage());
}
// *** End database connection




$username = ""; // Initialize value in order to keep its value so the user can still see it in his form

if (isset($_POST['login'])) { // if the "login" button is pressed
    $username = $_POST['username']; // retrieve username value from the form
    $password = $_POST['password']; // retrieve password value from the form

    /* 
    *   If a username is unique then a way to do it is to count how many times
    *  the couple with this username and this password appears in our database.
    */
    $query = $db->prepare("SELECT COUNT(*) userAmount ".
                            "FROM application_users ".
                            "WHERE username = $username ".
                            "AND password = $password;");

    $query->execute();
    $query->closeCursor();

    $resultAmount = $query->fetch();
    if ($resultAmount['userAmount'] == 0){ // If the couple username-password is unfound
        $message = "Username or password unknown";
    } else {
        $message("Login successful");
    }
}
?>


<h1>Login</h1>
<form action="login.php" method="post">
    Username:<br />
    <input type="text" name="username" placeholder="username" value="<?php echo($username); ?>" />
    <br/><br/>
    Password:<br/>
    <input type="password" name="password" placeholder="password" value="" />
    <br/><br/>
    <input type="submit" name="login" value="Login" />
<a href="register.php">Register</a>
</form>


登录
用户名：

您确定数据是通过post操作发送的吗？是否完成了任何基本调试？
var\u dump（$\u post）
和
var\u dump（$row）
将准确显示php从表单和数据库接收到的内容。检查字符串长度、检查字符串内容等。从未真正使用php进行过任何调试，我应该将var_dump语句放在何处。@Barry，我想是这样的，因为我正在学习一个教程，我知道在MYSQL数据库中密码是加密的，但在我的在我使用标准密码的情况下，我进入数据库并创建了另一个密码未加密的用户。您可以在任何地方添加密码，以查看参数中包含的变量。