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