2010-08-28 110 views
3

我有一個使用AJAX像這樣加載到我的網頁文章列表:jQuery:可以通過ajax加載內容更新DOM嗎?

var fm = <?php echo $from_user ;?>; 

$("#microblogposts").load("posts.php", {from_user: fm}, function(){ 
}); 

在帖子這個名單我有一個功能,從列表中刪除的帖子:

//START POST DELETE FUNCTION  
$("form#deletepost").submit(function() { 

// we want to store the values from the form input box, then send via ajax below 
var deleteid = $('#deleteid').attr('value'); 

//START AJAX  
$.ajax({ 
    type : "POST", 
    url: "process.php", 
    data: {deleteid : deleteid}, 
    error: function(){ 
    alert("Mesage could not be posted at this time. Please try again."); 
    }, 
    success: function(){ 

      $("#microblogposts").load("posts.php", {from_user: fm}, function(){ 
      //alert("posts have been loaded"); 
      }); 

      $("#latestpost").load("latestpost.php", {from_user: fm}, function(){ 
      //alert("latest posts have been loaded"); 
      }); 

    } 
    //END SUCCESS FUNCTION 
}); 
return false; 
//END AJAX 

}); 
//END POST DELETE FUNCTION 

我希望代碼在AJAX成功時刷新DOM中的某些元素,但它不起作用,有沒有反正可以這樣做?

+0

'.load(「...」,{...},function(d){console.log(d);})''的每個'.load()'函數的輸出是什麼? – jigfox 2010-08-28 17:00:12

+0

我該如何檢查? – user342391 2010-08-28 17:07:07

+0

change $(「#microblogposts」)。load(「posts.php」,{from_user:fm},function(){ // alert(「posts have been loaded」); });到$(「#microblogposts」)。load(「posts.php」,{from_user:fm},function(d){ console.log(d); }); ,看看控制檯說什麼。 – jigfox 2010-08-28 17:09:19

回答

1

感謝您的幫助jigfox。問題在於變量,它們都是空的,因此內容沒有加載。