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Asp.net mvc 鼠标位置ajax-asp.net MVC_Asp.net Mvc_Json_Serialization - Fatal编程技术网

Asp.net mvc 鼠标位置ajax-asp.net MVC

Asp.net mvc 鼠标位置ajax-asp.net MVC,asp.net-mvc,json,serialization,Asp.net Mvc,Json,Serialization,我使用以下方法将鼠标向下位置发送到控制器中的两个操作,第一个是动态生成图像,第二个是将图像发送到查看器: $("#container2").bind('mousemove', function (e) { currentX = e.pageX; currentY = e.pageY; diffX =

我使用以下方法将鼠标向下位置发送到控制器中的两个操作,第一个是动态生成图像,第二个是将图像发送到查看器:

   $("#container2").bind('mousemove', function (e) {
                                currentX = e.pageX;
                                currentY = e.pageY;
                                diffX = (currentX - startX);
                                diffY = (currentY - startY);
                                $.ajax({
                                    url: '/Home/Scale3D',
                                    type: 'POST',
                                    data: {
                                        pX: diffX,
                                        pY: diffY,
                                        pZ: diffX * sThickness,
                                        udm: ++udm
                                    }
                                }).done(function (d) {
                                    image.src = '/Home/Rotate3DImage?udm=' + (++udm);
                               });
                            });
以下是第一个行动:

 public ActionResult Scale3D(float pX = 0, float pY = 0, float pZ = 0, int inc = 0)
        {
            DicomImage3D im = null;
           ......
            return Json(new { success = true, inc = inc }, JsonRequestBehavior.AllowGet);
        }
第二项行动:

public ActionResult Rotate3DImage(int udm = 0)
    {
        FileContentResult data;
        ........
        using (var memStream = new MemoryStream())
        {
            objImage.Save(memStream, ImageFormat.Png);
            data = this.File(memStream.GetBuffer(), "image/png");
        }
        return data;
    }
问题是有一个延迟和一些图像不显示;直到第一个操作处理完所有第一个ajax命令,第二个操作才会执行。收到成功消息时,是否有方法执行第二个操作。非常感谢您的建议,提前谢谢。

谢谢。我改变了。完成(功能(d){成功。仍然有延迟。
$("#container2").bind('mousemove', function (e) {
                                currentX = e.pageX;
                                currentY = e.pageY;
                                diffX = (currentX - startX);
                                diffY = (currentY - startY);
                                $.ajax({
                                    url: '/Home/Scale3D',
                                    type: 'POST',
                                    data: {
                                        pX: diffX,
                                        pY: diffY,
                                        pZ: diffX * sThickness,
                                        udm: ++udm
                                    },
                                    success: {
                                     // code here
                                     image.src = '/Home/Rotate3DImage?udm=' + (++udm);
                                    }
                                });
                            });