Question

我想删除数据库中的行而不重新加载页面。我写了以下代码。它会删除页面中的行但不会删除数据库中的行。这意味着连接不会进入ajax。请帮助。

    <?php  


$advances_result= mysqli_query($conn, "SELECT * FROM advances") or die("error");
?>

        <table border>
            <caption>ADVANCES</caption>
        <tr>
            <th>Date</th>
            <th>EmpName</th>
            <th>EmpCode</th>
            <th>Amount</th>
            <th>Comment</th>
            <th>Action</th>
        </tr>

<?php
while($row = mysqli_fetch_array($advances_result)){
?>    
    <tr>
    <td><?php echo $row['date'];?></td>
    <td><?php echo $row['empname'];?></td>
    <td><?php echo $row['empcode'];?></td>
    <td style='text-align:right'><?php echo $row['amount'];?></td>
    <td><?php echo $row['comment'];?></td>
    <td><button class="delete" id="<?php echo $row['id']; ?>" > Delete</button></td>
    </tr>


<?php    
}
?>
        </table>

'jquery part'

                        <script>
                        $(document).ready(function(){                
                        $(".delete").click(function() {


                         if (confirm("Do you want to delete this row?"))
                                        {
                                            var row = $(this).parents('tr');
                                            var id = row.attr("id");
                                            var data = 'id=' + id ;

                        $.post({
                           type: "post",
                           url: "deleteadvances.php",
                            data: data,
                           cache: false,
                           success: function(){ 
                            row.slideUp('slow', function() {$(row).remove();});

                          }
                         });
                         }
                        return false;
                            });
                        });
                        </script>

'deleteadvances.php'页面：

<?php
include("connection.php");

$id = $_POST['id'];
mysqli_query($conn,"INSERT INTO advancehistory SELECT * FROM advances WHERE ID='".$id."'");
mysqli_query($conn,"DELETE FROM advances WHERE ID='".$id."'");



?>

Answer 1

你需要在deleteadvances.php中测试mysqli_query的输出，如果一切正常，它会显示JSON Ok消息，可以在jQuery端使用$ .post轻松解析。

例如：{"deleted":"id"}成功或{"error":404}未找到错误。

Answer 2

我认为您从javascript传递到ajax的ID不正确。

var data = 'id=' + id;

用字符串追加id＆＃39; id =＆＃39;现在它终于变成了一个字符串。当它传递给ajax时，只有一个值，而不是[key，value]对。所以，当你在php中访问它时，你会得到错误的id或Null值。

尝试：

$.post({
    type: "post",
    url: "deleteadvances.php",
    datatype : 'json',
    data: {
      id : id
    },
    cache: false,
    contentType : 'application/x-www-form-urlencoded; charset=utf-8'
});

Answer 3

更改' deleteadvances.php '页面：

<?php

    include ('connection.php');

    $id=$_POST['id'];
    $delete = "DELETE FROM table_name WHERE id=$id";
    $result = mysqli_query($db,$delete) or die("Bad SQL: $delete");

?>

表行在数据库中删除而不重新加载页面

3 个答案: