Question

我有以下代码，当你输入一个值时，var $ username不会回显。

//TODO: SET AUTH TOKEN as random hash, save in session
$auth_token =  rand();   

if (isset($_POST['action']) && $_POST['action'] == 'Login')
{

    $errors = array(); //USED TO BUILD UP ARRAY OF ERRORS WHICH ARE THEN ECHOED
    $username = $_POST['username'];

    if ($username = '')
    {
        $errors['username'] = 'Username is required';
    }

    echo $username; // var_dump($username) returns string 0
}


require_once 'login_form.html.php';

 ?>

login_form是这样的：

<form method="POST" action="">
<input type="hidden" name="auth_token" value="<?php echo $auth_token ?>">
Username: <input type="text" name="username">
Password: <input type="password" name="password1">
<input type="submit" name="action" value="Login">
</form>

身份验证令牌部分并不重要，只是当我在用户名文本框中输入值并按下登录按钮时，用户名不会回显，var_dump返回字符串（0），print_r只是空白。

Answer 1

愚蠢的错误

if ($username = '')   <-- this is an assignment

应该是这个

if ($username == '')  <-- this is comparison

Answer 2

if ($username = '')

您错过了=，因此您要为$username分配一个空字符串。将其更改为

if ($username == '')
               ^-- note the 2 equal signs.

Answer 3

此行是作业，而不是比较：

if ($username = '')

你想：

if ($username == '')

Answer 4

//TODO: SET AUTH TOKEN as random hash
$auth_token =  rand();   

if (isset($_POST['action']) && $_POST['action'] == 'Login')
{

    $errors = array(); //USED TO BUILD UP ARRAY OF ERRORS WHICH ARE THEN ECHOED
    $username = $_POST['username'];

    if ($username == '') // **You are assiging not comparing** 
    {
        $errors['username'] = 'Username is required';
    }

    echo $username; // var_dump($username) returns string 0
}


require_once 'login_form.html.php';

 ?>

在您的登录表单中:(操作属性..）

 <form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
 <input type="hidden" name="auth_token" value="<?php echo $auth_token ?>">
 Username: <input type="text" name="username">
 Password: <input type="password" name="password1">
 <input type="submit" name="action" value="Login">
 </form>

变量不会回声

4 个答案: