ajax成功后解析json


ajax post success parse json

很抱歉问了这么长的问题。我正在尝试ajax帖子来收集联系人位置历史记录,然后将标记添加到地图中。

ajax帖子返回位置数据json,编码如下:

[{"name":"Chris","data":{"user":"447967048843","data":[{"timestamp":1332840872,"longitude":-1.549517,"latitude":53.973174},{"timestamp":1332841510,"longitude":-1.444133,"latitude":53.997148}]},"contacts":null},{"name":"Jason","data":{"user":"447879896697","data":[{"timestamp":1332839836,"longitude":-1.566667,"latitude":53.978533},{"timestamp":1332840447,"longitude":-1.567654,"latitude":53.977927}]},"contacts":null}]

这里是getHistory函数,它在选择联系人后在表单提交时调用。

function getHistory() {
    var contact = $("#contact").val()
    var days = $("#days").val()
    $.ajax({
        type: 'post',
        url: 'temp_history.php',
        data: {contact: contact, days: days},
        context: document.body,
        beforeSend: function() {
        },
        success: function(succ){
            var obj = jQuery.parseJSON(succ);
            var divs="",tabs="",counts=0;
            jQuery("#gMap").gmap3({
                action: 'clear'});
                jQuery(".marker").remove();
                jQuery.each(obj,function(i,item){
                tabs +=item.contacts;
                if(item.data.latitude != null && item.data.longitude!=null)
                    {
                    addMarker(item.name, item.data.timestamp,item.data.latitude,item.data.longitude,item.data.user_id);
                    }
            });
        }
  });

}

我认为问题是我需要嵌套jQuery.each函数,但不确定如何嵌套?

addMarker函数为:

 function addMarker(name, timestamp, lati, longi, user_id) {
    jQuery("#gMap").gmap3({
        action: 'addMarkers',
        markers:[
        {lat:lati, lng:longi, data:name}
        ]
    }); 

}

感谢

你是对的-你的JSON遍历不正确,在你的success处理程序中试试这个:

success: function(data){
    var json = jQuery.parseJSON(data); // If you're returing json, this shouldn't be required
    var divs = "", tabs = "", counts = 0;
    jQuery("#gMap").gmap3({ action: 'clear' });
    jQuery(".marker").remove();
    jQuery.each(json, function(i, item) {
        var name = item.name;
        var userId = item.data.user;
        jQuery.each(item.data.data, function(i, nav) {
            var ts = nav.timestamp;
            var lat = nav.latitude;
            var long = nav.longitude;
            if (lat != null && long != null) {
                addMarker(name, ts, lat, long, userId);
            }
        });
    })  
}

此外,让JSON中的对象名称更具语义也是值得的,尤其是当data在多个级别中列出时。