getjson问题

Question

var items; 
$.getJSON('calender_service.php?command=calender', function(data) { 
    items = data; 
    var name = []; 
    var date_start = []; 

    var date_start_time = ""; 
    $.each(items, function(key, val) { 
     name[key] = val.name; 
     date_start[key] = val.date_start; 
     events[key] = ""+""+","+val.date_start.split("-")+","+val.date_start_time+","+val.date_end_time+","+val.name+","+val.status+""; 
    }); 

    alert(events); 
}); 

请帮助我，没有得到getjson方法外的变量值事件。

Answer 1

为了让它在回调函数之外，您必须将events变量定义为全局变量（即在进行json调用之前尝试定义它）。

Answer 2

试试这个：

var items; 
var events; 

$.getJSON('calender_service.php?command=calender', function(data) { 
    items = data; 
    var name = []; 
    var date_start = []; 
    var date_start_time = ""; 
    $.each(items, function(key, val) { 
     name[key] = val.name; 
     date_start[key] = val.date_start; 
     events[key] = "" + "" + "," + val.date_start.split("-") + "," + val.date_start_time + "," + val.date_end_time + "," + val.name + "," + val.status + ""; 
    }); 
}); 

alert(events); 

Answer 3

定义事件作为外部范围的变量。

var events = {}; 
var items; 
$.getJSON('calender_service.php?command=calender', function(data) { 
    items=data; 
var name=[]; 
    ... 
    .. 
}); 

Answer 4

定义$ .each循环之外的“events”。例如在date_start之后。