Question

你可以帮我解决这个问题吗？成功：....我得到了我需要的结果，但最后的ajaxFunction返回''值。

ajaxFunction = function(lastNumber){
        var json_result = '';
        $.ajax({
            type: "POST",
            contentType: "application/json",
            dataType: "json",
            data: '{"first" : "'+lastNumber+'"}',
            url: "http://localhost:8080/some_url",
            success: function (result) {
                console.log("in success:  " + result.result);
                json_result =  result.result;
            }
        });
        console.log("before return   " + json_result);
        return json_result;
    };

Answer 1

问题是返回结果，然后调用的ajax执行异步。

以下是使用Deferred替换的示例。

return value from nested function in jquery

function test(){
   var def = new $.Deferred();   
    $.ajax({
        url: 'test',
        success: function(data) {
            // Do some minor logic to determine if server response was success.
            console.log('success in callback');
            if(data == true){
              def.resolve(true);  
            }
            else{
              def.reject(); 
              // or possibly def.resolve(true) depending what you're looking for.
            }

        },
        error:function(){
          console.log('error in callback');
          def.reject();
        }
    });

  return def; 
}

编辑：您的案例：

＆＃13;

var ajaxFunction = function(lastNumber){
  var def = new $.Deferred();         
        $.ajax({
            type: "POST",
            contentType: "application/json",
            dataType: "json",
            data: '{"first" : "'+lastNumber+'"}',
            url: "http://example.com",
            success: function (result) {
                console.log("in success:  " + result.result);
                def.resolve(result.result);
            },
          error: function(){
            def.reject();
            }
        });
        return def;
    };

var testMethod = function(){
  var result = ajaxFunction('some number');
  result.done(function(result){
    alert(result);
    console.log('done in deferred');
    // Do something when everything went well.

  }).fail(function(){
    alert('error');
    console.log('error in deferred');
    // Do something when something went bad.
  });

}

testMethod();

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

Answer 2

你可以把它拿来试试

    ajaxFunction = function(lastNumber){
    var json_result = '';
    $.ajax({
        async: false,
        type: "POST",
        contentType: "application/json",
        dataType: "json",
        data: '{"first" : "'+lastNumber+'"}',
        url: "http://localhost:8080/some_url",
        success: function (result) {
            console.log("in success:  " + result.result);
            json_result =  result.result;
        }
    });
    console.log("before return   " + json_result);
    return json_result;
};

当您将async设置为false时，它会使调用同步，默认情况下为true。

如果不使用这个，你可以使用回叫或承诺（工作和返回，如同步调用）。

Answer 3

始终使用$.Deferred()对象并在Ajax调用中返回它。

在.resolve() Ajax调用回调中使用.success()到您的延迟变量。

调用您的方法并使用延迟函数的.done方法进行检查。

具有ajax请求的函数

3 个答案: