Estoy usando ImageCropper
y MediaPlugin
para Subir imágenes. Sin embargo, tengo problemas para conseguir la imagen después de recortar la imagen.
string imagefile;
protected void OnClickedRectangle(object sender, EventArgs e)
{
new ImageCropper()
{
Success = (imageFile) =>
{
Device.BeginInvokeOnMainThread(() =>
{
view_imageavatar.Source = ImageSource.FromFile(imageFile);
});
}
}.Show(this);
}
async void edit_avatar_Tapped(object sender, EventArgs e)
{
try
{
await CrossMedia.Current.Initialize();
new ImageCropper()
{
PageTitle = "Title",
AspectRatioX = 1,
AspectRatioY = 1,
CropShape = ImageCropper.CropShapeType.Rectangle,
SelectSourceTitle = "Img",
TakePhotoTitle = "Take Camera",
PhotoLibraryTitle = "Img Gallery",
Success = (imageFile) =>
{
Device.BeginInvokeOnMainThread(async() =>
{
view_imageavatar.Source = ImageSource.FromFile(imageFile);
imagefile = imageFile;
//API Get Images Upload
var content = new MultipartFormDataContent();
content.Add(new StreamContent(imageFile), "files", imagefile);
var httpClient = new HttpClient();
var responses = await httpClient.PostAsync("https://xxxxx/api/Upload", content);
});
}
}.Show(this);
}
catch (Exception ex)
{
System.Diagnostics.Debug.WriteLine("GalleryException:>" + ex);
}
}
Sin embargo ¿cómo puedo obtener la Imagen a cargar. tenga en cuenta que view_imageavatar
todavía muestra la imagen después de la cosecha. Tks!
Actualización de...
async void edit_avatar_Tapped(object sender, EventArgs e)
{
try
{
await CrossMedia.Current.Initialize();
new ImageCropper()
{
PageTitle = "Title",
AspectRatioX = 1,
AspectRatioY = 1,
CropShape = ImageCropper.CropShapeType.Rectangle,
SelectSourceTitle = "Img",
TakePhotoTitle = "Take Camera",
PhotoLibraryTitle = "Img Gallery",
Success = (imageFile) =>
{
Device.BeginInvokeOnMainThread(async() =>
{
view_imageavatar.Source = ImageSource.FromFile(imageFile);
imagefile = imageFile;
//API Get Images Upload
var fileStream = File.OpenRead(imageFile);
var fileContent = new StreamContent(fileStream);
var content = new MultipartFormDataContent();
content.Add(fileContent, "files", imageFile);
var httpClient = new HttpClient();
var responses = await httpClient.PostAsync("https://xxxxxx/api/UploadAvatarUs", content);
});
}
}.Show(this);
}
catch (Exception ex)
{
System.Diagnostics.Debug.WriteLine("GalleryException:>" + ex);
}
}
Aún así no funciona?
Actualización 2
async void edit_avatar_Tapped(object sender, EventArgs e)
{
try
{
await CrossMedia.Current.Initialize();
new ImageCropper()
{
PageTitle = "Title",
AspectRatioX = 1,
AspectRatioY = 1,
CropShape = ImageCropper.CropShapeType.Rectangle,
SelectSourceTitle = "Img",
TakePhotoTitle = "Take Camera",
PhotoLibraryTitle = "Img Gallery",
Success = (imageFile) =>
{
Device.BeginInvokeOnMainThread(async() =>
{
view_imageavatar.Source = ImageSource.FromFile(imageFile);
imagefile = imageFile;
//API Get Images Upload
var upfilebytes = File.ReadAllBytes(imageFile);
var ms = new MemoryStream(upfilebytes);
var content = new MultipartFormDataContent();
content.Add(new StreamContent(ms), "files", imageFile);
var httpClient = new HttpClient();
var responses = await httpClient.PostAsync("https://xxxxxx/api/UploadAvatarUs", content);
});
}
}.Show(this);
}
catch (Exception ex)
{
System.Diagnostics.Debug.WriteLine("GalleryException:>" + ex);
}
}
-> No puede subir fotos a través de la API?
Sin embargo, yo trate de no usar ImageCropper. Puedo subir directamente.
async void edit_avatar_Tapped(object sender, EventArgs e)
{
var file = await MediaPicker.PickPhotoAsync();
var content = new MultipartFormDataContent();
content.Add(new StreamContent(await file.OpenReadAsync()), "files", file.FileName);
var httpClient = new HttpClient();
var responses = await httpClient.PostAsync("https://xxxxxx/api/UploadAvatarUs", content);
string a = responses.StatusCode.ToString();
}
--> Entonces funciona bien, la imagen se carga a través de la API
¿La imagen de carga de content.Add(new StreamContent(ms), "files", imageFile);
no trabajar con la API? La búsqueda de soluciones por parte de todos.
File.Open(imageFile);
->Open
no funciona. Me paso enFile.OpenRead(imageFile)
. Es esto bueno? He actualizado el anterior.