我想做的只是使用mono touch / mono droid和mvvmcross将照片上传到网络服务,希望在某种程度上我只需要为android和IOS编写一次代码:)
我最初的想法是让用户选择一个图像(在android中使用意图)获取图像的路径。然后使用MvxResourceLoader resourceLoader从路径打开流,然后使用restsharp创建包含流的发布请求。
然而,我已经撞墙,当用户选择图像时,路径是例如“/外部/图像/媒体/ 13”。使用MvxResourceLoader resourceLoader时,此路径导致文件未找到异常。
为什么我会得到例外的任何想法,或者有更好的方法来实现我的目标?
答案 0 :(得分:3)
这就是我最终做到的方式 - 谢谢stuart和所有链接:)
public class PhotoService :IPhotoService, IMvxServiceConsumer<IMvxPictureChooserTask>,IMvxServiceConsumer<IAppSettings>
{
private const int MaxPixelDimension = 300;
private const int DefaultJpegQuality = 64;
public void ChoosePhotoForEventItem(string EventGalleryId, string ItemId)
{
this.GetService<IMvxPictureChooserTask>().ChoosePictureFromLibrary(
MaxPixelDimension,
DefaultJpegQuality,
delegate(Stream stream) { UploadImage(stream,EventGalleryId,ItemId); },
() => { /* cancel is ignored */ });
}
private void UploadImage(Stream stream, string EventGalleryId, string ItemId)
{
var settings = this.GetService<IAppSettings>();
string url = string.Format("{0}/EventGallery/image/{1}/{2}", settings.ServiceUrl, EventGalleryId, ItemId);
var uploadImageController = new UploadImageController(url);
uploadImageController.OnPhotoAvailableFromWebservice +=PhotoAvailableFromWebservice;
uploadImageController.UploadImage(stream,ItemId);
}
}
public class PhotoStreamEventArgs : EventArgs
{
public Stream PictureStream { get; set; }
public Action<string> OnSucessGettingPhotoFileName { get; set; }
public string URL { get; set; }
}
public class UploadImageController : BaseController, IMvxServiceConsumer<IMvxResourceLoader>, IMvxServiceConsumer<IErrorReporter>, IMvxServiceConsumer<IMvxSimpleFileStoreService>
{
public UploadImageController(string uri)
: base(uri)
{
}
public event EventHandler<PhotoStreamEventArgs> OnPhotoAvailableFromWebservice;
public void UploadImage(Stream stream, string name)
{
UploadImageStream(stream, name);
}
private void UploadImageStream(Stream obj, string name)
{
var request = new RestRequest(base.Uri, Method.POST);
request.AddFile("photo", ReadToEnd(obj), name + ".jpg", "image/pjpeg");
//calling server with restClient
var restClient = new RestClient();
try
{
this.ReportError("Billedet overføres", ErrorEventType.Warning);
restClient.ExecuteAsync(request, (response) =>
{
if (response.StatusCode == HttpStatusCode.OK)
{
//upload successfull
this.ReportError("Billedet blev overført", ErrorEventType.Warning);
if (OnPhotoAvailableFromWebservice != null)
{
this.OnPhotoAvailableFromWebservice(this, new PhotoStreamEventArgs() { URL = base.Uri });
}
}
else
{
//error ocured during upload
this.ReportError("Billedet kunne ikke overføres \n" + response.StatusDescription, ErrorEventType.Warning);
}
});
}
catch (Exception e)
{
this.ReportError("Upload completed succesfully...", ErrorEventType.Warning);
if (OnPhotoAvailableFromWebservice != null)
{
this.OnPhotoAvailableFromWebservice(this, new PhotoStreamEventArgs() { URL = url });
}
}
}
//method for converting stream to byte[]
public byte[] ReadToEnd(System.IO.Stream stream)
{
long originalPosition = stream.Position;
stream.Position = 0;
try
{
byte[] readBuffer = new byte[4096];
int totalBytesRead = 0;
int bytesRead;
while ((bytesRead = stream.Read(readBuffer, totalBytesRead, readBuffer.Length - totalBytesRead)) > 0)
{
totalBytesRead += bytesRead;
if (totalBytesRead == readBuffer.Length)
{
int nextByte = stream.ReadByte();
if (nextByte != -1)
{
byte[] temp = new byte[readBuffer.Length * 2];
Buffer.BlockCopy(readBuffer, 0, temp, 0, readBuffer.Length);
Buffer.SetByte(temp, totalBytesRead, (byte)nextByte);
readBuffer = temp;
totalBytesRead++;
}
}
}
byte[] buffer = readBuffer;
if (readBuffer.Length != totalBytesRead)
{
buffer = new byte[totalBytesRead];
Buffer.BlockCopy(readBuffer, 0, buffer, 0, totalBytesRead);
}
return buffer;
}
finally
{
stream.Position = originalPosition;
}
}
}
答案 1 :(得分:1)