Question

我正在为php分配编写一些代码，我从URL获得了正确的ID，该表显示了与该人相对应的所有正确记录，但是我的删除按钮无法正常工作，我要么删除了有关此人的表格，否则我会出错。

我的头部上方的PHP部分

<?php require "config/config.php"; ?>
<?php
    if(isset($_GET['upd'])){
        $id = $_GET['upd'];
        $query = "SELECT * FROM persons WHERE id=$id";
        $fire = mysqli_query($con,$query) or die("Can not fetch the data.".mysqli_error($con));
        $user = mysqli_fetch_assoc($fire);
        }
?>

我在头部上方的删除部分

<?php   
        if(isset($_GET['delweight'])){
            $weightid = ($_GET['weightid']);
            $query = "DELETE FROM personweight WHERE weightid = $weightid";
            $fire = mysqli_query($con,$query) or die("Can not delete the data from database.". mysqli_error($con));

            if($fire) echo "Data deleted from database";

            }   
?>

带有删除记录的我的表

<table class="table table-striped table-dark" id="weightTable">
        <thead>
            <tr><th>weightid</th><th>Weight</th><th>Date</th><th>Delete</th></tr>
        </thead>
        <tbody>
            <?php       
                $query = "SELECT * FROM personweight WHERE id=$id";
                    $fire = mysqli_query($con,$query) or die("can not fetch data from datase ".mysqli_error($con));
                        if(mysqli_num_rows($fire)>0){
                        while($user = mysqli_fetch_assoc($fire)){ ?>
            </tr>
                <td><?php echo $user['weightid'] ?></td>
                <td><?php echo $user['weight'] ?></td>
                <td><?php echo $user['added'] ?></td>
                <td>
                    <a href="<?php $_SERVER['PHP_SELF'] ?>?delweight=<?php echo $user['id'] ?>" class="btn btn-sm btn-danger">Delete</a>
                </td>
            </tr>
                <?php }}  ?>
        </tbody>
    </table>

PHP删除具有多个未联接表的记录

0 个答案: