使用jquery/ajax加载php页面以打开特定链接以在Div中显示页面

我有一个带有id/name/desc字段的mysql表，我使用php选择查询在div中显示它

<div class="show_name">
    while( $row = $data->fetch_array(MYSQLI_ASSOC))
    { ?>
        <div><?php echo $row['id'];?></div>
        <div><?php echo $row['name'];?></div>
    <?php } ?>
</div>
<div id="content"></div>

我有一个name.php页面，我想在其中显示单击的名称链接的描述

for example
id name
1   abc
2   xyz

如果我点击abc name链接，我应该能够用abc desc打开name.php

如果我单击xyz名称链接，我应该能够用xyz desc打开相同的name.php

等等

---

请给予任何帮助，谢谢

放置用于存储id的类和用户定义的数据属性

// add clickMe class
// and add data-id attribute
<div class="clickMe" data-id="<?php echo $row['id'];?>"><?php echo $row['id'];?></div>
<div><?php echo $row['name'];?></div>

在

name.php

中应该有如下查询和html页面：

// this id we pass from load request earlier 
$myId = $_POST['id'];
// do your logic here
// query from database for details description using where id = '$myId'
// display the content here
// suppose you have data store inside $myData variable
$myData = getData();
echo $myData['name'];
echo $myData['desc'];

// using on to delegate for dynamic element
$('.show_name').on('click','.clickMe', function(){
  // capture data id
  var id = $(this).data('id');
  // load name.php and pass post parameter(id)
  // we pass id paramater here
  // you can use success callback if want to populate anything
  $('#content').load('name.php', {id : id});
  // above code should display data from name.php
});

// this id we pass from load request earlier 
$myId = $_POST['id'];
// do your logic here
// query from database for details description using where id = '$myId'
// display the content here
// suppose you have data store inside $myData variable
$myData = getData();
echo $myData['name'];
echo $myData['desc'];