相关疑难解决方法(0)

这应该返回一个包含图片文件名列表的JSON对象.注释警报显示正确的数据,但alert(getPicsInFolder("testfolder"));显示"error".

function getPicsInFolder(folder) {
  return_data = "error";
  $.get("getpics.php?folder=" + folder, function (data) {
    data = jQuery.parseJSON(data);
    $.each(data, function (index, value) {
      data[index] = "folders/" + folder + "/" + value;
    });
    //alert(data); // This alert shows the correct data, but that's hardly helpful
    return_data = data;
  });
  return return_data;
}

我究竟做错了什么？

我将数据从JQuery ajax调用传递回调用位置时出现问题.有问题的代码如下:

jQuery("#button").click(function()
{   
    for(var i = 0;i < data.length; i++)
    {
        result = updateUser(data[i]); //result is not populated..
                    alert(result); //prints 'undefined'

    }
});

function updateUser(user_id)
{       
    jQuery.ajax({
        url:"/users/update/"+user_id,
        type:"GET",
        async: false,
        success: (function(data){           
            //if I alert "data" here it shows up correctly
                            //but if i try to return it like below 
                            //it does not get passed correctly 
                            return data; 
        })
    });

任何指针都非常感谢