如何获取 HTML 和 PHP 中的输入值

Question

我正在尝试创建更新模式，但我无法从名为 update_player_id 的隐藏输入中获取player_id。

我尝试使用$_GET和$_POST，但没有成功。我将输入从隐藏更改为文本和player_id，它已正确传递。

$_GET 和 $_POST 的 var_dump 是 Array(0) { }，如果我尝试使 $player_id = $_GET['update_player_id'] 我有以下输出：警告：未定义的数组键“Update_player_id”，与 $ 相同玩家id = $_POST['update_player_id']

<div id="update-player-modal" class="update-player-modal">
    <div class="modal-content">
        <span class="close" onclick="closeUpdatePlayerModal()">&times;</span>
        <h2>Update Player</h2>
        <div class="form">
            <form action="process-update-player.php" method="POST">
                <input type="hidden" name="update_player_id" value="">                

                <?php
                echo $_GET['update_player_id'];
                $player_id = $_GET['update_player_id'];
                                        
                $mysqli = require "../database/database.php"; 
                $sql = "SELECT * FROM player WHERE id = ?";
                $stmt = $mysqli->prepare($sql);
                    
                $stmt->bind_param("i", $player_id);
                    
                $stmt->execute();
                    
                $result = $stmt->get_result();
                    
                if ($result->num_rows > 0) {
                    
                    while ($row = $result->fetch_assoc()) {
                        echo "<div class='fullname'>";
                        echo "<div>";
                        echo "<label for='firstName'>First Name</label>";
                        echo "<input type='text' id='firstName' name='firstName' value='" . $row["firstName"] . "'>";
                        echo "</div>";
                        echo "<div>";
                        echo "<label for='lastName'>Last Name</label>";
                        echo "<input type='text' id='lastName' name='lastName' value='" . $row["lastName"] . "'>";
                        echo "</div>";
                        echo "</div>";
                            
                        echo "<div>";
                        echo "<label for='birthday'>Birthday</label>";
                        echo "<input type='text' id='birthday' name='birthday' value='" . $row["birthday"] . "'>";
                        echo "</div>";
                            
                        echo "<div>";
                        echo "<label for='nationality'>Nationality</label>";
                        echo "<input type='text' id='nationality' name='nationality' value='" . $row["nationality"] . "'>";
                        echo "</div>";
                            
                        echo "<div>";
                        echo "<label for='alias'>Alias</label>";
                        echo "<input type='text' id='alias' name='alias' value='" . $row["alias"] . "'>";
                        echo "</div>";
                            
                        echo "<div>";
                        echo "<label for='position'>Position</label>";
                        echo "<input type='text' id='position' name='position' value='" . $row["position"] . "'>";
                        echo "</div>";
                            
                        echo "<div>";
                        echo "<label for='team'>Team</label>";
                        echo "<input type='text' id='team' name='team' value='" . $row["team"] . "'>";
                        echo "</div>";
                    }
                } else {
                    echo "Player not found!";
                }
                    
                $stmt->close();
                $mysqli->close();
                ?>
                <div class="update-player-button">
                    <button>Update</button>
                </div>
            </form>
        </div>     
    </div>
</div>

Answer 1

完成

David评论后，您可能会在执行任何表单请求之前尝试阅读update_player_id

。

首先，如果您使用

<input type="hidden" name="update_player_id" value="">

提交，您将不会有任何数据，因为属性

name

为空。

因此，您需要先获取

player id

，无论它来自其他页面、按钮等，因为您无法从上面的输入中获取它，然后您可以在显示表单并填写之前搜索

player

它与玩家信息。

<?php
    // ...
    
    // This value could get from the url http://....?update_player_id=4
    $player_id = $_GET['update_player_id']; // -> 4
                                        
    $mysqli = require "../database/database.php"; 
    $sql = "SELECT * FROM player WHERE id = ?";
    $stmt = $mysqli->prepare($sql);

    // ...

最后，您可以使用

heredoc 使字段生成更清晰，而无需 echo

 语句，并且可以选择使用 

array destructuring

 代替 

$row["key"]

。