I'm looking for a way to send model and image in one request. I tried to send my model from body, but I don't know how to send the file. Is there another way than sending image and model in a different files?
Here's the POST method from my API:
[HttpPost]
[Route("UploadNewEvent")]
public async Task<IActionResult> CreateNewEventAsync([FromBody] EventModel model)
{
var file = this.Request.Form.Files.LastOrDefault();
if (file != null)
{
var uploads = Path.Combine(_environment.WebRootPath, "uploads");
using (var fileStream = new FileStream(Path.Combine(uploads, "test.jpg"), FileMode.Create))
{
await file.CopyToAsync(fileStream);
}
}
// do sth with model later
return Ok();
}
And here's the code from my App:
public async Task SendNewEvent(EventModel model, MediaFile photo)
{
var uri = $"{baseUri}api/User/Event/CreateNewEvent";
if (photo != null)
{
var multipartContent = new MultipartFormDataContent();
multipartContent.Add(new StreamContent(photo.GetStream()), "\"file\"", $"\"{photo.Path}\"");
var httpClient = new HttpClient();
var jsonObject = JsonConvert.SerializeObject(model);
var stringContent = new StringContent(jsonObject, Encoding.UTF8, "application/json");
var httpResponseMessage = await httpClient.PostAsync(uri, stringContent);
}
}
For passing Model with File parameter, you need to post data as form-data.
Follow steps below:
Change FromBody
to FromForm
[HttpPost]
[Route("UploadNewEvent")]
public async Task<IActionResult> CreateNewEventAsync([FromForm] EventModel model)
{
// do sth with model later
return Ok();
}
Change client code to send form-data instead of json string
var uri = $"https://localhost:44339/UploadNewEvent";
FileStream fileStream = new FileStream(@"filepath\T1.PNG", FileMode.Open);
var multipartContent = new MultipartFormDataContent();
multipartContent.Add(new StreamContent(fileStream), "\"file\"", @"filepath\T1.PNG");
// EventModel other fields
multipartContent.Add(new StringContent("2"), "Id");
multipartContent.Add(new StringContent("Tom"), "Name");
var httpClient = new HttpClient();
var httpResponseMessage = httpClient.PostAsync(uri, multipartContent).Result;
EventModel
public class EventModel
{
public int Id { get; set; }
public string Name { get; set; }
public IFormFile File { get; set; }
}