Question

为什么我的isset和/或空函数不起作用？每次我单击html表单上的提交按钮时，即使表单为空，它也会插入到数据库中。有人可以让我彻底解释我做错了什么以及如何解决它？

<?php 
$host = "localhost";
$user = "root";
$db_password = "";
$database = "listings_db";

$link = mysqli_connect($host, $user, $db_password, $database) or die("Error " . mysqli_error($link));

      if (isset
     ($_POST['user_firstname'], 
      $_POST['user_lastname'], 
      $_POST['user_email'], 
      $_POST['user_password'], 
     $_POST['user_type'])) 
{
$firstname = $_POST['user_firstname'];
$lastname = $_POST['user_lastname'];
$email = $_POST['user_email'];
$password = $_POST['user_password'];
$type = $_POST['user_type'];

$errors = array();
if(empty($firstname) 
|| empty($lastname) 
|| empty($email) 
|| empty($email) 
|| empty($password) 
|| empty($type)) 
{$errors [] = '*All fields are required!';}     
else {
if(filter_var($email, FILTER_VALIDATE_EMAIL) === false) {$errors[] = '*Please enter a valid email address!' ;}

if(strlen($firstname) > 25) {$errors[] = '*The email address you entered contains too many characters!';}
if(strlen($lastname) > 25) {$errors[] = '*The first name you entered contains too many characters!';}
if(strlen($email) > 40) {$errors[] = '*The last name you entered contains too many characters!';}   
if(strlen($password) > 40) {$errors[] = '*The password you entered contains too many characters!';} 
if(strlen($type) != true){$errors[] = '*Please select an account type!';}
    }

$firstname = $_POST['user_firstname'];
$lastname = $_POST['user_lastname'];
$email = $_POST['user_email'];
$password = md5($_POST['user_password']);
$type = $_POST['user_type'];

$firstname = mysqli_real_escape_string($link, $firstname);
$lastname  = mysqli_real_escape_string($link, $lastname);
$email = mysqli_real_escape_string($link, $email);
$password = mysqli_real_escape_string($link, $password);
$type = mysqli_real_escape_string($link, $type);


$query = mysqli_query($link, "INSERT INTO users (user_id, user_firstname, user_lastname, user_email, user_password) VALUES ('', '$firstname', '$lastname', '$email', '$password')");
}
?>

编辑：以下是表格：

<form action="" method="post">
<p>First Name: <br><input type="text" name="user_firstname" size="25" maxlength="25"/></p>
<p>Last Name: <br><input type="text" name="user_lastname" size="25" maxlength="25"/></p>
<p>Email Address: <br><input type="email" id="email" name="user_email" size="25" maxlength="40"/><p>
<p>Create a Password: <br><input type="password" name="user_password" size="25" maxlength="40"/></p>
<p>Account Type: <br> <select name="user_type"> 
<option value="seller" selected>Seller</option>
<option value="Buyer">Buyer </option>
</select>

Answer 1

这是因为您的表单字段将始终提交。无论他们是否充满了任何内容。

因此，您无需检查它们是否已设置，因为它们将始终设置。如果它们不是空的，你需要检查。为此，您可以使用空的PHP构造：

if (!empty($_POST['user_firstname'])
    && !empty($_POST['user_lastname'])
    && !empty($_POST['user_email'])
    && !empty($_POST['user_password'])
    && !empty($_POST['user_type']))
{
    // TODO
}

Isset和空函数不起作用

1 个答案: