範圍問題：值丟失

Question

我想從一個方法（pollServiceForInfo）返回一個JSON對象，但是當我在方法完成後提醒它時它似乎會「丟失」。我知道這是一個範圍界定問題，但是我很難理解如何進行。洞察力將不勝感激。

var id=null; 
var jsonData = JSON.stringify({searchRequest:{coordinates: "1,2,3 1,2,3 1,2,3 1,2,3 1,2,3"}}); 
$.post("rest/search",jsonData, function(json){ 
    id = json.searchResponse.id; 
}) 
.error(function(jqXHR, textStatus, errorThrown){ 
    alert("obj.responseText: "+jqXHR.responseText + " textStatus: "+textStatus+" errorThrown: "+errorThrown); 
}) 
.success(function(data, status, obj){ 
    // process initial request 
    var json = pollServiceForInfo(id); // method below 

    alert(json); // says undefined 
}); 



var pollServiceForInfo = function(id){ 
    //alert('id in pollServiceForInfo '+id);  
    var jsonResults; 
    $.get("rest/poll/"+id,function(data){ 
     jsonResults = data.pollResponse; 

    }).error(function(){ 
     alert('returning error'); 
     return "error"; 
    }).success(function(){ 
     alert('returning data '+jsonResults); 
     return jsonResults; // is lost after it's returned 
    }); 
}; 

Answer 1

你不能有用地從異步函數返回。請改爲：

var pollServiceForInfo = function(id, callback){ 
    //alert('id in pollServiceForInfo '+id);  
    var jsonResults; 
    $.get("rest/poll/"+id,function(data){ 
     jsonResults = data.pollResponse; 

    }).error(function(){ 
     alert('returning error'); 
     callback("error"); 
    }).success(function(){ 
     alert('returning data '+jsonResults); 
     callback(jsonResults); // is lost after it's returned 
    }); 
}; 

pollServiceForInfo(id, function(json) { 
    alert(json); 
}); 

Answer 2

您試圖從成功回調中返回。你想要的是內pollServiceForInfo（）的返回，就像這樣：

var pollServiceForInfo = function(id){  
    var jsonResults; 
    $.get("rest/poll/"+id,function(data){ 
     jsonResults = data.pollResponse; 
    }).error(function(){ 
     alert('returning error'); 
     jsonResults = "error"; 
    }).success(function(){ 
     alert('returning data '+jsonResults);   
    }); 

    return jsonResults; 
};