Question

我目前正在我的学院学习PHP，这是我很困惑的事情。我不知道如何使用HTML表单在PHP中创建下拉菜单以从数据库中删除行。这是我的代码：

//This is index.php
    <div>
        <?php
            include('includes/delete.php');
        ?>
    <form method="POST" action="includes/delete.php">

        <select name="username">
        <?php
            $sql = mysqli_query($connection, "SELECT user_name, user_lastname FROM users");
            while ($row = mysqli_fetch_assoc($sql)){?>
                <option>
                <?php echo $row['user_name'];?>
                <?php echo $row['user_lastname'];?>
                </option> 
            <?php }?>
            </select>
        <input type="submit" name="delete" value="Delete">
    </form>
    </div>

//This is delete.php
<?php

    $serverName = "localhost";
    $userName = "root";
    $userPassword = "";
    $dbName = "baza_podataka";

    $connection = mysqli_connect($serverName, $userName, $userPassword, $dbName);

    //Checking connection
    if (!$connection) {
        die("Error: Could not connect!" . mysqli_connect_error());
    }

    //Deleting an user from database
    if(isset($_POST['delete'])) {
    $username = $_POST['username'];

        if(mysqli_query($connection, "DELETE FROM 'users' WHERE 'username' = '$username' " )){
        echo "User was deleted!";
        }else{
            echo "User could not be deleted! Please try again later.";
        }
    }

Answer 1

delete是按钮的ID，不会发布。您需要检查$ _post ['username']。

if（isset（$ _ POST ['username']））{ $ username = $ _POST ['username']; ...

Answer 2

代码中缺少的是option标记内缺少select值。同时，您的sql代码容易受到sql注入攻击，因此您应该使用bind参数。好的，您将使用以下代码片段来运行代码。

  <option value="<?php echo $row['user_name']?>" >
      <?php echo $row['user_name'];?>
      <?php echo $row['user_lastname'];?>
  </option>

如何创建一个使用PHP从数据库中删除行的下拉菜单？

2 个答案: