Question

是否可以将javascript中的msg变量从回调函数中返回？我尝试这样做，但得到一个null，即使msg变量在回调函数的范围内有数据。

var msg = load();


function load()
{
     $.ajax({  
     type: "POST",  
     url: "myPage.aspx/MyMethod",  
     data: jsonText,  
     contentType: "application/json; charset=utf-8",  
     dataType: "json",  
     success: function (msg) { return msg;  // Doesn't return anything because
                                            //it's automatically called on success??? 
     },  
     failure: function () { alert("Failure"); }  
     }); 
}

Answer 1

不是异步请求，这是正常的ajax请求和首选类型。那是因为load在您收到服务器的回复之前返回，所以很明显它无法返回msg。相反，让load接受回调：

load(function(msg) {
    // Use msg here
});

function load(callback)
{
    $.ajax({  
        type: "POST",  
        url: "myPage.aspx/MyMethod",  
        data: jsonText,  
        contentType: "application/json; charset=utf-8",  
        dataType: "json",  
        success: function (msg) {
            // Call our callback with the message
            callback(msg);
        },  
        failure: function () {
            // Call our callback with an indication things failed
            callback(null); // Or however you want to flag failure
        }
     }); 
}

如果绝对不可避免，您可以通过在$.ajax选项中设置async: false（这是文档的链接）来使用同步请求，然后执行以下操作：

var msg = load();

function load(callback)
{
    var result;
    $.ajax({  
        type: "POST",  
        url: "myPage.aspx/MyMethod",  
        data: jsonText,  
        contentType: "application/json; charset=utf-8",  
        dataType: "json",  
        async: false,
        success: function (msg) {
            // Set the `result` value here. We can't *return* it
            // here because, of course, that would just return it
            // from our `success` function.
            result = msg;
        },  
        failure: function () {
            result = null; // Or however you want to flag failure
        }
     });

     // Because we've made the request synchronous, we won't get here
     // until the ajax call is complete and `result` has been set by
     // the callbacks above; we can now return it.
     return result;
}

但同步请求会导致糟糕的用户体验，而且几乎不需要，所以应该尽可能避免。

Answer 2

您需要将回调传递给load()，因为XHR是异步的。

BTW，如果您没有明确地返回任何内容，则msg应包含undefined。

从回调函数中返回JSON数据

2 个答案: