Question

我正在尝试创建一个表来管理上传的音乐，我希望当用户按下“删除”时它会删除该条目。这是我正在使用的代码。

<?php
$sql="SELECT * FROM content WHERE `uploader` = '" . $user_data['username'] . "' ORDER BY id DESC";
$records=mysql_query($sql);
while($sound=mysql_fetch_assoc($records)){
    echo "<tr class='adder'>";
    echo "<td width='40' class='player'>&nbsp;&nbsp;<a href='".$sound['link']."' class='sm2_button'>Play/</a></td>";
    echo '<td width="75" class="name">'.$sound['date'].'</td>';
    echo '<td width="150" class="name">'.$sound['name'].'</td>';
    echo "<td width='58' class='bpm'>".$sound['uploader']."</td>";
    echo "<td width='220' class='keywords'>".$sound['keywords']."</td>";
    echo "<td width='50' class='keywords'>Edit</td>";
    echo "<td width='50' class='keywords'><span onclick='mysql_query('DELETE FROM content WHERE id = ".$sound['id']." ')>Delete</span></td>";
    echo "</tr>";
    }
?>

Answer 1

我不是php开发人员，但我相信你应该将数据提交给服务器。看来您的代码只是打印出来的＆＃39; mysql_query（＆＃39; DELETE FROM content WHERE id =＆＃34;。$ sound [＆＃39; id＆＃39;]。＆＃34;＆＃39;）

到html页面。

你应该做的是

通过POST或GET示例（deleteSound.php？soundId = 1）将数据提交给服务器
deleteSound.php然后将使用php代码（服务器代码）删除id = 1的声音
删除声音后，将其重定向回原始页面（soundlist.php）

Answer 2

这真的很简单

创建一个名为process.php的页面

<?php
// include your connection file

$id = "0";
if (isset($_GET['id'])) {
  $id = $_GET['id'];
}

try {
    $sql = "DELETE FROM table-name WHERE id=:id";
    $stmt = $conn->prepare($sql);
    $stmt->bindValue(':id', $id, PDO::PARAM_INT);
    $stmt->execute();
} catch (PDOException $e) {
    die("Could not delete from the table: " . $e->getMessage());
}
  $GoTo = "/the/path/where/you/want/to/send/the/visitor.php";
  header(sprintf("Location: %s", $GoTo));
?>

然后改变这一行......

echo "<td width='50' class='keywords'><span onclick='mysql_query('DELETE FROM content WHERE id = ".$sound['id']." ')>Delete</span></td>";

为...

echo "<td width='50' class='keywords'><a href=\"process.php?id=".$sound['id']."\">Delete</a></td>";

希望这有帮助。

快乐的编码！

Answer 3

我只是重新使用一个更安全的方法是使用$ _POST ...使用$ _GET任何用户都可以通过更改'process.php删除其他用户条目？id = x'

我必须使用表单按钮才能执行此操作....

表格行

echo "<td width='50' class='keywords'><form name='deletesound' class='deletesound' method='post' action='process.php'>

“;

<强> process.php：

<?php
ob_start();
include 'core/init.php'; 
include 'includes/header.php';
protect_page();

$id = $_POST['getid'];
if (isset($_POST['getid'])) {
  $id = $_POST['getid'];
  mysql_query("DELETE FROM content WHERE id = $id");
  header('Location: manage.php');
  exit();
}
?>

Answer 4

使用ajax尝试以下代码。

viewContent.php

<?php
$sql="SELECT * FROM content WHERE `uploader` = '" . $user_data['username'] . "' ORDER BY id DESC";

$records=mysql_query($sql);
while($sound=mysql_fetch_assoc($records)){
?>
  <tr class='adder'>
    <td width='40' class='player'>&nbsp;&nbsp;<a href='<?php echo $sound['link']; ?>' class='sm2_button'>Play/</a></td>
    <td width="75" class="name"><?php echo $sound['date']; ?></td>
    <td width="150" class="name"><?php echo $sound['name']; ?></td>
    <td width='58' class='bpm'><?php echo $sound['uploader']; ?></td>
    <td width='220' class='keywords'><?php echo $sound['keywords']; ?></td>
    <td width='50' class='keywords'>Edit</td>
    <td width='50' class='keywords'><span onclick='deleteContent(<?php echo $sound['id']; ?>)'>Delete</span></td>
    </tr>
    }
?>

<script>
function deleteContent(id)
{
var xmlhttp;
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    window.location="/viewContent.php";
    }
  }
xmlhttp.open("GET","deleteContent.php?q="+id,true);
xmlhttp.send();
}
</script>

deleteContent.php

<?php
$id=$_GET['id'];
mysql_query('DELETE FROM content WHERE id = ".$id." ')
echo "done";

?>

上面的代码从视图页面发送请求以通过ajax删除页面，并在获取repsonse时重新加载页面。

单击元素时，Mysql会删除表中的条目

4 个答案: