为什么从表单输入没有通过

Question

我是新来的php，并创建了两种形式。一种形式接收用户的名字，另一种形式接收用户的姓氏。我不知道为什么当用户输入文本到表单没有被传递。

在PHP与形式

<?php include '../view/header.php'; ?> 
<div id="main"> 
    <h1>Add Athlete</h1> 
    <form action="index.php" method="post" id="add_product_form"> 
     <input type="hidden" name="action" value="add_product" /> 

     <label>Country:</label> 
     <select name="category_id"> 
     <?php foreach ($categories as $category) : ?> 
      <option value="<?php echo $category['categoryID']; ?>"> 
       <?php echo $category['categoryName']; ?> 
      </option> 
     <?php endforeach; ?> 
    </select> 
    <br /> 

    <label>Code:</label> 
    <input type="input" name="code" /> 
    <br /> 

    <label>First Name:</label> 
    <input type="input" name="first_name" /> 
    <br /> 

    <label>Last Name:</label> 
    <input type="input" name="last_name" /> 
    <br /> 

    <label>&nbsp;</label> 
    <input type="submit" value="Add Athlete" /> 
    <br /> <br /> 
    </form> 
    <p><a href="index.php?action=list_products">View List of Olympic Athletes</a></p> 

</div> 
<?php include '../view/footer.php'; ?> 

，在输入采用从窗体

<?php 
require('../model/database.php'); 
require('../model/product_db.php'); 
require('../model/category_db.php'); 

if (isset($_POST['action'])) { 
    $action = $_POST['action']; 
} else if (isset($_GET['action'])) { 
    $action = $_GET['action']; 
} else { 
    $action = 'list_products'; 
} 

if ($action == 'list_products') { 
    // Get the current category ID 
    $category_id = $_GET['category_id']; 
    if (!isset($category_id)) { 
     $category_id = 1; 
} 

// Get product and category data 
$category_name = get_category_name($category_id); 
$categories = get_categories(); 
$products = get_products_by_category($category_id); 

    // Display the product list 
    include('product_list.php'); 
} else if ($action == 'delete_product') { 
    // Get the IDs 
    $product_id = $_POST['product_id']; 
    $category_id = $_POST['category_id']; 

    // Delete the product 
    delete_product($product_id); 

    // Display the Product List page for the current category 
    header("Location: .?category_id=$category_id"); 
} else if ($action == 'show_add_form') { 
    $categories = get_categories(); 
    include('product_add.php'); 
} else if ($action == 'add_product') { 
    $category_id = $_POST['category_id']; 
    $first_name = ""; 
    if(isset($_POST['first_name'])){$first_name = $_POST['FirstName'];} 
    $last_name = ""; 
    if(isset($_POST['last_name'])){$last_name = $_POST['LastName'];} 



    // Validate the inputs 
    if (empty($first_name) || empty($last_name)) { 
     $error = "Invalid product data. Check all fields and try again."; 
     include('../errors/error.php'); 
    } else { 
     add_product($category_id, $first_name, $last_name); 

     // Display the Product List page for the current category 
     header("Location: .?category_id=$category_id"); 
    } 
} else if ($action == 'list_categories') { 
    $categories = get_categories(); 
    include('category_list.php'); 
} else if ($action == 'add_category') { 
    $first_name = $_POST['FirstName']; 

    // Validate inputs 
    if (empty($name)) { 
     $error = "Invalid category name. Check name and try again."; 
     include('view/error.php'); 
    } else { 
     add_category($name); 
     header('Location: .?action=list_categories'); // display the Category List page 
    } 
} else if ($action == 'delete_category') { 
    $category_id = $_POST['category_id']; 
    delete_category($category_id); 
    header('Location: .?action=list_categories');  // display the Category List page 
} 
?> 

的add_product功能

function add_product($category_id, $first_name, $last_name) { 
    global $db; 
    $query = "INSERT INTO products 
       (categoryID, FirstName, LastName) 
       VALUES 
       ('$category_id', '$first_name', '$last_name')"; 
    $db->exec($query); 
} 
?> 

Answer 1

除了@ MahfuzulAlam的回答，另一个问题是：

if(isset($_POST['first_name'])){$first_name = $_POST['FirstName'];} 
                 ^^^^^^^^^ 

if(isset($_POST['last_name'])){$last_name = $_POST['LastName'];} 
                ^^^^^^^^ 

它应该是：

if(isset($_POST['first_name'])){$first_name = $_POST['first_name'];} 

if(isset($_POST['last_name'])){$last_name = $_POST['last_name'];} 

此外，在PHP 7，还有一个速记你目前在做什么：

$first_name = $_POST['first_name'] ?? ""; 

$last_name = $_POST['last_name'] ?? ""; 

它叫做Null Coalesce Operator。

Answer 2

尝试使用类型= text不input PHP的。 <input type="text" name="last_name" />等。

而且这个代码有问题。

if(isset($_POST['first_name'])){$first_name = $_POST['FirstName'];} 
if(isset($_POST['last_name'])){$first_name = $_POST['LastName'];} 

无论你通过为FIRST_NAME和last_name您$first_name和$last_name变量将是空的。因为您正在使用$_POST['FirstName'], $_POST['LastName']来分别设置它们的值，所以这不是真的存在并且都返回NULL。改写如下这些行：

if(isset($_POST['first_name'])){$first_name = $_POST['first_name'];} 
if(isset($_POST['last_name'])){$first_name = $_POST['last_name'];}