Ajax 调用访问 Web 服务器，但没有执行回调函数

Question

ajax函数

function Verify(ccode,dgh)
{
    str = "ccode="+ccode+"&dgh="+dgh;
    console.log(str);//this outputs means that this functions gets called
    $.ajax({
        type: "POST",
        url: "ajax/verify",
        data: str,
        async: false,
        cache: false,
        error: function (xhr, ajaxOptions, thrownError)
        {
            console.log(xhr.status);
            console.log(thrownError);                         
        },
        success: function(json)
        {
            console.log("in-fun: "+json.code); //does not gets executed
            return json.code; //does not return value
        },
        failure:function(response)
        {
            console.log("Ajax call failed"); //does not executes
        }
    });
}

上面的ajax函数被称为var e = Verify(var1, var2); e 的值在 ajax 请求后未定义。

ajax 请求确实到达了我的 Web 服务器，并且在 apache 日志和开发工具中可见，并返回 200 OK。 Ajax 端点正在工作并且确实返回了一个有效的 json。页面输出头也设置为json

编辑：更新了上面的代码

function Verify(ccode,dgh)
{
    var retData = '';
    str = "ccode="+ccode+"&dgh="+dgh;
    console.log(str); // this works
    $.ajax({
        type: "POST",
        url: "ajax/verify",
        data: str,
        async: false,
        cache: false,
        error: function (xhr, ajaxOptions, thrownError)
        {
            console.log(xhr.status); //does not gets called
            console.log(thrownError);

        },
        success: function(json)
        {
            console.log("in-fun: "+json.code); //this does not ouputs anything
            retData = json.code;
        },
        complete:function(response)
        {
            console.log("Complete called"); //does not gets called
        }
    });
    return retData;
}

Answer 1

由于已经使用了 jQuery AJAX 调用，您可以依赖它的 deferred object，如下所示：

function Verify(ccode, dgh)
{
  var str = "ccode="+ccode+"&dgh="+dgh;
  console.log(str);  //debug outputs
  return $.ajax({
      type: "POST",
      url: "ajax/verify",
      data: str
   });
}

Verify(var1, var2).done(function(json) {
  if (json) {
    var e = json.code;
    // more code for the success case
  }
  else {
    console.log("Invalid server response"); 
  }
}).fail(function() {
  console.log("Ajax call failed"); 
});