如何从服务器获取照片并将其显示在Image Control for Wp7中

时间:2011-03-20 00:43:21

标签: c# windows-phone-7

我正在使用以下代码。我只是不知道它为什么不起作用。错误消息是:未指定错误:bmp.SetSource(ms)。

我不熟悉Wp7的HttpWebRequest。非常感谢您帮助解决这个问题。感谢。

enter code here


 private void LoadPic()
    {
   HttpWebRequest req = (HttpWebRequest)HttpWebRequest.Create(@"http://xxxxxx/MyImage.jpg");
        NetworkCredential creds = new NetworkCredential("Username", "Pwd");
        req.Credentials = creds;
        req.Method = "GET";
        req.BeginGetResponse(new AsyncCallback(GetStatusesCallBack), req);
    }

    public void GetStatusesCallBack(IAsyncResult result)
    {
        try
        {
            HttpWebRequest httpReq = (HttpWebRequest)result.AsyncState;
            HttpWebResponse response = (HttpWebResponse)httpReq.EndGetResponse(result);
            Stream myStream = response.GetResponseStream();
            int len = (int)myStream.Length;

            byte[] byt = new Byte[len];
            myStream.Read(byt, 0, len);
            myStream.Close();
            MemoryStream ms = new MemoryStream(byt);
            Deployment.Current.Dispatcher.BeginInvoke(() =>
            {
                BitmapImage bmp = new BitmapImage();
                bmp.SetSource(ms);

                image1.Source = bmp;
            });
        }
        catch (Exception ex)
        {
            MessageBox.Show(ex.Message);
        }
   }

1 个答案:

答案 0 :(得分:1)

是否有必要将响应流复制到字节数组,然后复制到MemoryStream?如果没有,您可以执行以下操作:

    Stream myStream = response.GetResponseStream();
    Deployment.Current.Dispatcher.BeginInvoke(() => {
        BitmapImage bmp = new BitmapImage();
        bmp.SetSource(myStream);
        image1.Source = bmp;
    });

如果由于某种原因必须复制,则需要在循环中填充缓冲区:

    Stream myStream = response.GetResponseStream();
    int contentLength = (int)myStream.Length;
    byte[] byt = new Byte[contentLength];
    for (int pos = 0; pos < contentLength; )
    {
        int len = myStream.Read(byt, pos, contentLength - pos);
        if (len == 0)
        {
            throw new Exception("Upload aborted.");
        }
        pos += len;
    }
    MemoryStream ms = new MemoryStream(byt);
    Deployment.Current.Dispatcher.BeginInvoke(() =>
    {
        // same as above
    });

第二部分(稍微)改编自C# bitmap images, byte arrays and streams!