【问题标题】:Expect json array in AJAX期望 AJAX 中的 json 数组
【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+"
        &notification_targetuser="+notification_targetuser+
       "&notification_triggeredby="+notification_triggeredby,   
       dataType:"json",
       success: function(data){
       //do something 
       )

【问题讨论】:

    标签: php arrays json ajax while-loop


    【解决方案1】:

    使用json.parse()

    success: function(data){
         var a = JSON.parse(data);
         //loop var a to get your data 
    }
    

    查看this 链接,您可能会有所了解。

    【讨论】:

    • 应该是 data.notification_id
    • a 现在是一个数组。应该是a[0].notification_id
    • 谢谢,正是我想要的。
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