Question

我正在努力用我的下拉菜单删除数据库中的数据。我的下拉菜单看起来像这样

<form method="post" action="admin.php">
<h3>Delete a user</h3>
<select name="username">
$sql = mysqli_query($connection, "SELECT username FROM users");
while ($row = $sql->fetch_assoc()){
?>

<option value="username" name="username">
<?php echo $row['username']; ?></option>
<?php } ?>
<input type="submit" name="delete" value="Delete User">
</form>

这显示用户都很喜欢我想要它，所以这里是php for it

<?php 
include('connect.php');
if(isset($_POST['delete'])) {
$username = $_POST['username'];
mysqli_query("DELETE FROM `users` WHERE `username` = '$username' ");
echo "User was deleted!";
}
?>

因此，当我点击提交按钮＆＃34;删除用户＆＃34;时，看起来我被发送到admin.php并且没有任何反应。

我该如何解决这个问题？感谢。

Answer 1

将name="username"替换为<option></option>
选项value中的回声值。
admin.php 页面

更新代码

<select name="username">
  <?php
  $sql = mysqli_query($connection, "SELECT username FROM users");
  while ($row = $sql->fetch_assoc()){?>
    <option value="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></option>
  <?php }?>
</select>

<强> admin.php的

mysqli_query($connection, "DELETE FROM `users` WHERE `username` = '$username' ");

Answer 2

您没有将值放在选择选项中，这就是为什么没有发生任何事情只需替换

<option value="username" name="username">
<?php echo $row['username']; ?></option>
<?php } ?>

带

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

它适用于你。

Answer 3

更改

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

要

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

Answer 4

1. <option value="username" name="username">需要<option value="<?php echo $row['username']; ?>">

2.缺少连接变量。需要： -

mysqli_query($connection,"DELETE FROM `users` WHERE `username` = '$username' ");

修改后的代码必须是： -

表格代码： -

<?php
//comment these two lines when code started working fine
error_reporting(E_ALL);
ini_set('display_errors',1);

include('connect.php');
?>
<form method="post" action="admin.php">
    <h3>Delete a user</h3>
    <select name="username">
      <?php
      $sql = mysqli_query($connection, "SELECT username FROM users");
      while ($row = mysqli_fetch_assoc($sql)){?>
        <option value="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></option>
      <?php }?>
    </select>
<input type="submit" name="delete" value="Delete User">
</form>

Php代码： -

<?php
//comment these two lines when code started working fine
error_reporting(E_ALL);
ini_set('display_errors',1);

include('connect.php');
if(isset($_POST['delete'])) {
    $username = $_POST['username'];
    if(mysqli_query($connection,"DELETE FROM `users` WHERE `username` = '$username' ")){
        echo "User was deleted!";
    }
}
?>

注意： - 总是做一些error-reporting，这样你就会收到错误并纠正错误。

您的查询容易受SQL INJECTION影响，因此请阅读prepared statements并使用它们。

使用PHP中的下拉菜单删除数据库中的数据

4 个答案: