我用下面的代碼獲取使用AJAX,jQuery的,PHP,JSON成員及其相關信息的列表。 唯一的問題是,當我使用.html,它只顯示第一個記錄,它不顯示所有的記錄。完全困惑。如何使用的.html中顯示多個記錄
<script type="text/javascript">
$(document).delegate("#member_home", "pagecreate", function() {
var refreshId = setInterval(function(){
var friends= new Array();
$.ajaxSetup({cache: false})
$.ajax({
url:'http://www.l.example.com/app/scrip/friends_lookup.php',
data: "",
dataType: 'json',
success: function(data){
$.each(data, function(key, val) {
var friend = val['friend'];
var phone = val['phone'];
var status = val['status'];
var email = val['email'];
var updated = val['updated'];
$('#member_friends').append("<div class='member-box'>"+friend+"<span class='status-pic1'><img src='images/"+status+".png' width='40' height='40'/></span><span class='phone_box'><a href='tel:"+phone+"'><img src='images/icons/phone.png' width='40' height='40' /></a></span><span class='email-box'><a href='mailto:"+email+"'><img src='images/mail.png' width='40' height='40' /></a></span><div class='clear'></div><span class='update-status'><i>last update: "+updated+"</i></span>");
});
}
});
}, 1500);
});
</script>
我想這一點,並沒有奏效:
<script type="text/javascript">
$(document).delegate("#member_home", "pagecreate", function() {
var refreshId = setInterval(function() {
var friends= new Array();
$.ajaxSetup({cache: false})
$.ajax({
url: 'http://www.l.example.com/pp/scripts/friends_lookup.php',
data: "",
dataType: 'json',
success: function(data){
var output = [];
for (var i = 0, len = data.length; i < len; i++) {
output[output.length] = {
friend : data[i].friend,
phone : data[i].phone,
status : data[i].status,
email : data[i].email,
updated : data[i].updated
};
$('#member_friends').html("<div class='member-box'>"+friend+"<span class='status-pic1'><img src='images/"+status+".png' width='40' height='40'/></span><span class='phone_box'><a href='tel:"+phone+"'><img src='images/icons/phone.png' width='40' height='40' /></a></span><span class='email-box'><a href='mailto:"+email+"'><img src='images/mail.png' width='40' height='40' /></a></span><div class='clear'></div><span class='update-status'><i>last update: "+updated+"</i></span>");
}
}
});
}, 1500);
});
</script>
與您的代碼縮進請告訴我?很難讀... – elclanrs 2012-02-09 00:15:56
我很抱歉,我從我的編輯複製和由於某些原因,它被認爲是搞砸了的 – 2012-02-09 00:16:54
可能重複[出於某種原因,我不能讓我的jQuery代碼的工作(http://stackoverflow.com /問題/ 9204052 /換一些,原因-I-着,讓 - 我 - jQuery的代碼到工作) – 2012-02-09 01:28:57