在php中向用户显示错误消息的正确方法是什么?

时间:2013-03-06 13:45:18

标签: php

我有一个简单的登录系统,但是当他们没有正确输入他们的凭据时,我希望能够将消息放在<p></p>或类似的东西中,而不是回显错误消息。我尝试使用空数组

$data = array();

然后在出现错误时设置状态变量而不是使用echo

$data["status"] = "...."

并在我放入的html部分

<?php if ( isset($status) ) : ?>
    <p><?= $status; ?></p>
<?php endif; ?>

但我似乎无法让它发挥作用。我也尝试过:

extract($data)

但是我显然做错了。我不确定这是不是正确的方法。使用javascript或jquery是一种下注方式来处理这个问题吗?无论如何这里是我的代码。

$conn = DBconnect($config);
//create and empty array so we can ifor the users.
$data = array();

// if the user has submitted the form
if( $_SERVER["REQUEST_METHOD"] === "POST") {

    //protect the posted value then store them to variables
    $username = protect($_POST["username"]);
    $password = protect($_POST["password"]);

    //Check if the username or password boxes were not filled in
    if ( !$username || !$password ){
        // if not display an error message.
        $data["status"] =  "You need to fill in a username and password!";
    }else{
        //check if the username and password match.
        $stmt = query("SELECT * FROM users WHERE username = :username AND password = :password",
                        array("username" => $username, "password" => $password),
                        $conn);
        $row = $stmt->fetch(PDO::FETCH_ASSOC);
        if(!$row){
            //display an error message
            $data["status"] =  "Username or Password is incorrect.";
        }elseif($row["active"] !=1) {
            $data["status"] =  "You have not activated your account!";
            }else{
                //we log the user in.

继续前进。这里是html部分

<div class="container">

  <form class="form-signin" action ="" method = "post">
    <h2 class="form-signin-heading">Please Log In</h2>
    <input type="text" class="input-block-level" name = "username" id = "username" placeholder="Username">
    <input type="password" class="input-block-level" name = "password" id = "password" placeholder="Password">
    <label class="checkbox">
      <input type="checkbox" value="remember-me"> Remember me
    </label>
     <input type="submit" class = "btn btn-large btn-primary" name = "submit" value="Log in!" >

    <?php if ( isset($status) ) : ?>
      <p><?= $status; ?></p>
    <?php endif; ?>

  </form>

</div> <!-- /container -->

1 个答案:

答案 0 :(得分:1)

$data["status"] = "...."
and on the html part i put in

<?php if ( isset($status) ) : ?>
    <p><?= $status; ?></p>
<?php endif; ?>

据我所知,$status未设置。 $data["status"]已设置。

根据评论中的建议,传递错误消息的正确方法;因为此错误是一个仅在当前会话中真正相关的值,所以在$_SESSION中传递它是有意义的。而且因为我们希望为所有值提供相关标题,将其存储在$_SESSION['error'];中是传递错误信息的完全合法方式。