【发布时间】:2016-02-07 14:13:50
【问题描述】:
我怎么可能让我的 Ajax 在我的成功中期待我的阵列。目前我假设我使用一个平面对象,这需要获取所有传递回客户端的数据,以便我可以更新我的通知。
我认为其他一切都是正确的!
{"num":1,"670":{"notification_id":"670","notification_content":"
Lucy Botham posted a status on your
wall","notification_throughurl":"singlepoststreamitem.php?
streamitem_id=545","notification_triggeredby"
"85","notification_status":"1"},"671":
{"notification_id":"671","notification_content":"Lucy Botham
posted a status on your
wall","notification_throughurl":"singlepoststreamitem.php?streamitem_id=546"
,"notification_triggeredby":"85","notification_status":"1"}}
服务器端
while($row = mysqli_fetch_assoc($com)){
$id = $row['notification_id'];
$num = mysqli_num_rows($com);
if($num){
$json['num'] = 1;
}else{
$json['num'] = 0;
}
$json[$id]['notification_id'] = $row['notification_id'];
$json[$id]['notification_content'] = $row['notification_content'];
$json[$id]['notification_throughurl'] = $row['notification_throughurl'];
$json[$id]['notification_triggeredby'] = $row['notification_triggeredby'];
$json[$id]['notification_status'] = $row['notification_status'];
}
echo json_encode($json);
客户
function loadIt() {
$.ajax({
type: "GET",
url: "viewajax.php?
notification_id="+notification_id+"
¬ification_targetuser="+notification_targetuser+
"¬ification_triggeredby="+notification_triggeredby,
dataType:"json",
success: function(data){
//do something
)
【问题讨论】:
标签: php arrays json ajax while-loop