Question

这是我的Login.php

<?php

//load and connect to MySQL database stuff
require("config.inc.php");

if (!empty($_POST)) {

    if(empty($_POST['username']) || empty($_POST['password'])) {

        $response["success"] = 0;
        $response["message"] = "Please fill in the login details!";
        die(json_encode($response));
    }

    $query = "SELECT  email, password, position FROM user   WHERE   email = :email ";

    $query_params = array(':email' => $_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;
    $login_ok = false;

    //fetching all the rows from the query
    $row = $stmt->fetch();
    if ($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!";
            $response["posts"]   = array();

                foreach ($row as $rerow) {
                $row = array(
                $post["position"] = $rerow["position"]
                            );

                array_push($response["posts"], $post);
                }

                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>
    <?php
}
?>

我不确定哪个地方我没有放近位。你能帮我看看吗？我不知道在哪里放近距离。在将我的数据检索到我的android时，是否与我的状况有关？如果是，请说出我知道我写的语法是否正确。

Answer 1

缺少if (!empty($_POST)) {

的}

    if (!empty($_POST)) {


     //code...

    } 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>
    <?php
    }
} // <---- this is missing

Answer 2

如果可以的话，我想告诉你，你的代码有点混乱。压痕可以挽救你的生命，让人记住。

无论如何，如果您删除代码的第五行行，我发现没有错误：

if (!empty($_POST)) {

这未关闭。 if ($row) {在登录标题标题上方的else之前关闭。事实上，如果数据库没有返回任何结果，它会显示登录页面。

无论如何，第五行是不必要的，因为在它下方你可以控制特定的变量。您不必检查完整的$_POST数组。

解析错误：语法错误，第95行的C：\ xampp \ htdocs \ PMSS \ login.php中的意外文件结束

2 个答案: