Is there any better way to work with ajax,php and mysql than this?

Question

Let's suppose that I want to have the following small code

<div id="field"> Some links here with a database call. </div>

The problem is that I want to be able to work with ajax on them and call them when the user adds a new link through the appropriate form.

So instead of having this div, I use a function and call it whenever needed using ajax.

     <?php  
  function testfield ($id) {  Database call
     ?>
      <div id="field"> <?php The links in here  ?> </div>
      }
   <html>
      <body>
        <?php testfield($_GET['id']); ?> 
     </body>
   </html>

So when a user adds a new link, I just call with ajax the function and I'm set.

Is there any other better way to do this? Thank you in advance

Answer 1

For example you should use JSON and return data into your div. For AJAX try use jQuery.

PHP code:

$linkarray = array();
$linkarray[] = 'link1';
$linkarray[] = 'link2';
$linkarray[] = 'link3';
echo json_encode($linkarray);

HTML:

<html>
<body><div id="linklist"></div></body>
</html>

JS:

$.ajax({
 url:'getlinklist.php',
        dataType:'json',
        type:'POST',
        success: function(data){
            $('#linklist').html('');
            $.each(datamfunction(key,value){
                $('#linklist').append(value+'<br/>');                 
            });
        }
    });

Answer 2

当用户添加新链接时，您将使用AJAX将数据发送到与数据库一起使用的PHP脚本，如果成功，则应使用javascript刷新DIV-在DIV末尾写入新信息（您已经拥有此信息），或使用信息，您从JSON大量获取，从PHP脚本成功返回

Is there any better way to work with ajax,php and mysql than this?

Question

2 answers

solution1
0 2011-11-07 09:54:05

solution2
0 2011-11-07 10:07:15

Is there any better way to work with ajax,php and mysql than this?

Question

2 answers

solution1 0 2011-11-07 09:54:05

solution2 0 2011-11-07 10:07:15

solution1
0 2011-11-07 09:54:05

solution2
0 2011-11-07 10:07:15