Question

我是Ajax的新手，我正在尝试使用相同的CRUD操作。我有一个 index.php 页面，我可以插入和显示数据，我也有链接来更新和删除数据.Create，Read，Update工作正常。现在的问题是，当我点击删除链接时，我已成功将ID发送到 delete_back.php ，并且该值已被删除但成功消息应该打印在< strong> index.php 因为我在那边有一个名为 #delete_msg 的div，但不知何故成功消息正在 delete_back.php上打印

的index.php

<div id="delete_msg"></div>

select_back.php

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

$data=mysqli_query($con,"select * from student");
$col=mysqli_num_fields($data);
echo "<table>";
while($row=mysqli_fetch_array($data))
{
    echo "<tr>";
    for($i=0;$i<$col;$i++)
    {
        echo "<td>".$row[$i]."</td>";   
    }   
    echo "<td><a href='update_front.php?id=$row[0]&nm=$row[1]&add=$row[2]&cont=$row[3]'> Update </a></td>";
    echo "<td><a class='delete' href='delete_back.php?id=$row[0]'>Delete</a></td>";
    echo "</tr>";
}
echo "</table>";
?>

delete_back.php

<?php
$id=$_GET['id'];

include("config.php");
$data=mysqli_query($con,"delete from student where id='$id'");
if($data=="true")
{
    echo "Data Deleted";    
}
else
{
    echo "Delete Error";    
}
?>

ajax文件

$(".delete").click(function(event){
        event.preventDefault();
        $.ajax({
            url:"delete_back.php",
            dataType:"html",
            success:function(msgStr){
                $("#delete_msg").html(msgStr);
        }
    })
  })

Answer 1

将data-id属性添加到锚标记：

mystr = 'student1 90\nstudent2 85\nstudent3 98'

[[i[0], int(i[1])] for i in (k.split() for k in mystr.split('\n'))]

# [['student1', 90], ['student2', 85], ['student3', 98]]

你可以在你的ajax中做到：

echo "<td><a id='delete' data-id='".$row[0]."' href='#'>Delete</a></td>";

在你的php文件中：

$("#delete").click(function(event){
        var obj = $(this); // first store $(this) in obj
        var id = $(this).data('id'); // get id of data using this 
        $.ajax({
            url:"delete_back.php",
            dataType:"html",
            data: { id: id },
            type: "Post",
            success:function(msgStr){
                $("#delete_msg").html(msgStr);

        }
    })
  })

Answer 2

首先，我真的希望你没有一个打印多个具有相同ID的锚点的循环...... 也就是说，根据你的HTML，你的JS应该像跟随，否则你将失去你的id值：

$("#delete").click(function(event){
        event.preventDefault();
        var my_href = $(this).attr('href');
        $.ajax({
            url: my_href,
            dataType:"html",
            success:function(msgStr){
                $("#delete_msg").html(msgStr);

            }
        });
        return false;
  });

但这不是一个非常好的编码风格..我建议使用html5 data-属性来存储您的id值

echo "<td><a id='delete' data-id='".$row[0]."' href='#'>Delete</a></td>";

如果您不想更改网址，也可以使用javascript:;代替#。

你的JS应该像：

$("#delete").click(function(event){
        // you can avoid preventing default behaviour because your anchor doesn't have any href value
        var del_id= $(this).data('id');
        $.ajax({
            url: "delete_back.php",
            dataType:"html",
            data: {id: del_id} // use object for better reading
            success:function(msgStr){
                $("#delete_msg").html(msgStr);
            }
        });
  });

我还建议您查看.done()构建LINK和deferred.done()

Answer 3

我认为删除链接来自包含其他删除链接的循环。如果是这样，这就是你应该做的。

echo '<td><a class="lnk-delete" href="#" data-id="'.$row[0].'">Delete</a></td>'; //The delete links should have 1 class name

然后进行ajax调用

$('.lnk-delete').click(function(e){
    var id = $(this).data('id');
    ajax({
        url: 'delete_back.php'
        data: {id: id},
        type: 'POST',
        success: function(data){
           $('#delete_msg').html(data);
        }
    });
});

获取要在php中删除的项目的ID

$id = $_POST['id'];

Answer 4

with from_to_table as (
    SELECT tid, valid_from,  LEAD(valid_from, 1, null) OVER (ORDER BY 
    valid_from)-1 AS valid_to,fee
    FROM   fee_table
    )
    select fee from from_to_table
    where to_date(:mydate,'dd/mm/rrrr')  between valid_from and nvl(valid_to,to_date(:mydate,'dd/mm/rrrr')+1)

成功消息显示在错误的页面上

的index.php

select_back.php

delete_back.php

ajax文件

4 个答案: