Question

当我尝试删除按钮时，没有错误，页面刷新，但是数据不会被删除，所以我认为问题在于从行中传递ID。

到目前为止，没有其他解决方案对我有用。这是桌子的身体。

<tbody>
<?php

$server = "localhost";
$user = "Website";
$pass = "pass";
$db = "db";

$conn = new mysqli($server, $user, $pass, $db);

if ($conn->connect_error)
{
    die("connection to database failed");
}

$query = "SELECT * FROM koppeling";
$result = mysqli_query($conn, $query);

while($row = $result-> fetch_assoc())
{
    echo "<tr>";
    echo "<td>".$row['Wagon_ID']."</td>";
    echo "<td name='wagon'>".$row['EPC']."
    <a href='delete.php' class='table-button' 
    id='".$row['id']."'>Delete</a>";
    echo "<td>";
    echo "</td>";
    echo "</tr>"; 
}
    ?>
</tbody>

delete.php文件：

<?php
$id = $_GET['id'];
$server = "localhost";
$user = "Website";
$pass = "pass";
$db = "db";

$conn = new mysqli($server, $user, $pass, $db);

if ($conn->connect_error)
{
    echo "Conn db failed";``
}

$query = "DELETE FROM koppeling WHERE id='$id';";

try
{
    mysqli_query($conn, $query);
    mysqli_close($conn);
    header('Location: koppelen.php');
}

catch (Exception $e)
{
    echo "$e";
}
?>

任何人都可以帮助我找出问题所在吗，我已经坚持了很长时间。

Answer 1

如果不是这样

while($row = $result-> fetch_assoc())
{
    echo "<tr>";
    echo "<td>".$row['Wagon_ID']."</td>";
    echo "<td name='wagon'>".$row['EPC']."
    <a href='delete.php' class='table-button' 
    id='".$row['id']."'>Delete</a>";
    echo "<td>";
    echo "</td>";
    echo "</tr>"; 
}

将此

while($row = $result-> fetch_assoc())
{
    echo "<tr>";
    echo "<td>".$row['Wagon_ID']."</td>";
    echo "<td name='wagon'>".$row['EPC']."
    <a href='delete.php?id=".$row['id']."' class='table-button' 
    id='".$row['id']."'>Delete</a>";
    echo "<td>";
    echo "</td>";
    echo "</tr>"; 
}

Answer 2

如果您将表单与POST请求或ajax一起使用，而不是使用链接，那会更好。

如果您真的想用这种方法，可以将查询字符串添加到href属性。

tmp.Color = xlRange.Cells[i, j].Interior.Color.ToString();

Answer 3

您必须更改以下行

$ query =“从koppeling那里删除，其中id ='$ id';”;

到

$ query =“从koppeling删除，其中id ='”。$ id。“';;;;

以便在查询中正确填充ID值

表格行删除按钮

3 个答案: