Return file in ASP.Net Core Web API
Here is a simplistic example of streaming a file:
using System.IO;
using Microsoft.AspNetCore.Mvc;
[HttpGet("{id}")]
public async Task<FileStreamResult> Download(int id)
{
var path = "<Get the file path using the ID>";
var stream = File.OpenRead(path);
return new FileStreamResult(stream, "application/octet-stream");
}
Note:
Be sure to use FileStreamResult
from Microsoft.AspNetCore.Mvc
and not from System.Web.Mvc
.
If this is ASP.net-Core then you are mixing web API versions. Have the action return a derived IActionResult
because in your current code the framework is treating HttpResponseMessage
as a model.
[Route("api/[controller]")]
public class DownloadController : Controller {
//GET api/download/12345abc
[HttpGet("{id}")]
public async Task<IActionResult> Download(string id) {
Stream stream = await {{__get_stream_based_on_id_here__}}
if(stream == null)
return NotFound(); // returns a NotFoundResult with Status404NotFound response.
return File(stream, "application/octet-stream"); // returns a FileStreamResult
}
}
Note:
The framework will dispose of the stream used in this case when the response is completed. If a
using
statement is used, the stream will be disposed before the response has been sent and result in an exception or corrupt response.
You can return FileResult with this methods:
1: Return FileStreamResult
[HttpGet("get-file-stream/{id}"]
public async Task<FileStreamResult> DownloadAsync(string id)
{
var fileName="myfileName.txt";
var mimeType="application/....";
Stream stream = await GetFileStreamById(id);
return new FileStreamResult(stream, mimeType)
{
FileDownloadName = fileName
};
}
2: Return FileContentResult
[HttpGet("get-file-content/{id}"]
public async Task<FileContentResult> DownloadAsync(string id)
{
var fileName="myfileName.txt";
var mimeType="application/....";
byte[] fileBytes = await GetFileBytesById(id);
return new FileContentResult(fileBytes, mimeType)
{
FileDownloadName = fileName
};
}