加載PHP頁面打開特定鏈接在一個div

Question

我有mysql表ID /名稱/遞減領域.I'm使用PHP選擇查詢

<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> 

在DIV顯示它顯示的頁面的jQuery/Ajax代碼

$('.show_name').on('click',function(){ 
    $('#content').load('name.php'); 
}); 

我name.php頁面，在這裏我想顯示點擊名稱鏈接的描述

for example 
id name 
1 abc 
2 xyz 

，如果我在ABC名稱鏈接點擊我應該能夠與ABC遞減

打開name.php如果我在某某名稱鏈接點擊我應該能夠打開同一name.php與XYZ遞減

請任何幫助appricated謝謝！

Answer 1

認沽類和用戶定義的數據屬性用於存儲ID

// 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> 

然後你的js代碼應該是這樣的：

// 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 
}); 

而且裏面你name.php應該有查詢和HTML頁面，如：

// 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'];