如何使用 Web API 返回文件？

Question

I am using ASP.NET Web API .
I want to download a PDF with C# from the API (that the API generates).

Can I just have the API return a byte[] ? and for the C# application can I just do:

byte[] pdf = client.DownloadData("urlToAPI");? 

and

File.WriteAllBytes()?

Answer 1

Better to return HttpResponseMessage with StreamContent inside of it.

Here is example:

public HttpResponseMessage GetFile(string id)
{
    if (String.IsNullOrEmpty(id))
        return Request.CreateResponse(HttpStatusCode.BadRequest);

    string fileName;
    string localFilePath;
    int fileSize;

    localFilePath = getFileFromID(id, out fileName, out fileSize);

    HttpResponseMessage response = new HttpResponseMessage(HttpStatusCode.OK);
    response.Content = new StreamContent(new FileStream(localFilePath, FileMode.Open, FileAccess.Read));
    response.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
    response.Content.Headers.ContentDisposition.FileName = fileName;
    response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/pdf");

    return response;
}

UPD from comment by patridge : Should anyone else get here looking to send out a response from a byte array instead of an actual file, you're going to want to use new ByteArrayContent(someData) instead of StreamContent (see here ).

Answer 2

I made the follow action:

[HttpGet]
[Route("api/DownloadPdfFile/{id}")]
public HttpResponseMessage DownloadPdfFile(long id)
{
    HttpResponseMessage result = null;
    try
    {
        SQL.File file = db.Files.Where(b => b.ID == id).SingleOrDefault();

        if (file == null)
        {
            result = Request.CreateResponse(HttpStatusCode.Gone);
        }
        else
        {
            // sendo file to client
            byte[] bytes = Convert.FromBase64String(file.pdfBase64);


            result = Request.CreateResponse(HttpStatusCode.OK);
            result.Content = new ByteArrayContent(bytes);
            result.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
            result.Content.Headers.ContentDisposition.FileName = file.name + ".pdf";
        }

        return result;
    }
    catch (Exception ex)
    {
        return Request.CreateResponse(HttpStatusCode.Gone);
    }
}

Answer 3

Just a note for .Net Core : We can use the FileContentResult and set the contentType to application/octet-stream if we want to send the raw bytes. Example:

[HttpGet("{id}")]
public IActionResult GetDocumentBytes(int id)
{
    byte[] byteArray = GetDocumentByteArray(id);
    return new FileContentResult(byteArray, "application/octet-stream");
}

Answer 4

Example with IHttpActionResult in ApiController .

[HttpGet]
[Route("file/{id}/")]
public IHttpActionResult GetFileForCustomer(int id)
{
    if (id == 0)
      return BadRequest();

    var file = GetFile(id);

    IHttpActionResult response;
    HttpResponseMessage responseMsg = new HttpResponseMessage(HttpStatusCode.OK);
    responseMsg.Content = new ByteArrayContent(file.SomeData);
    responseMsg.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
    responseMsg.Content.Headers.ContentDisposition.FileName = file.FileName;
    responseMsg.Content.Headers.ContentType = new MediaTypeHeaderValue("application/pdf");
    response = ResponseMessage(responseMsg);
    return response;
}

If you don't want to download the PDF and use a browsers built in PDF viewer instead remove the following two lines:

responseMsg.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
responseMsg.Content.Headers.ContentDisposition.FileName = file.FileName;

Answer 5

I've been wondering if there was a simple way to download a file in a more ... "generic" way. I came up with this.

It's a simple ActionResult that will allow you to download a file from a controller call that returns an IHttpActionResult . The file is stored in the byte[] Content . You can turn it into a stream if needs be.

I used this to return files stored in a database's varbinary column.

    public class FileHttpActionResult : IHttpActionResult
    {
        public HttpRequestMessage Request { get; set; }

        public string FileName { get; set; }
        public string MediaType { get; set; }
        public HttpStatusCode StatusCode { get; set; }

        public byte[] Content { get; set; }

        public Task<HttpResponseMessage> ExecuteAsync(CancellationToken cancellationToken)
        {
            HttpResponseMessage response = new HttpResponseMessage(StatusCode);

            response.StatusCode = StatusCode;
            response.Content = new StreamContent(new MemoryStream(Content));
            response.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment");
            response.Content.Headers.ContentDisposition.FileName = FileName;
            response.Content.Headers.ContentType = new MediaTypeHeaderValue(MediaType);

            return Task.FromResult(response);
        }
    }

Answer 6

Another way to download file is to write the stream content to the response's body directly:

[HttpGet("pdfstream/{id}")]
public async Task  GetFile(long id)
{        
    var stream = GetStream(id);
    Response.StatusCode = (int)HttpStatusCode.OK;
    Response.Headers.Add( HeaderNames.ContentDisposition, $"attachment; filename=\"{Guid.NewGuid()}.pdf\"" );
    Response.Headers.Add( HeaderNames.ContentType, "application/pdf"  );            
    await stream.CopyToAsync(Response.Body);
    await Response.Body.FlushAsync();           
}

Answer 7

You Can try , HttpClient for Download file from another side and same time you can pass as File Result

 [HttpGet]
    [Route("api/getFile")]
    public async  Task<FileResult> GetFile(string Param1,string Param2)
    {
        try
        {
            Stream stream = null;
            string strURL = @"File URL";
            HttpClient client = new HttpClient();
            HttpResponseMessage httpResponse = await client.GetAsync(strURL);
            Stream streamToReadFrom = await httpResponse.Content.ReadAsStreamAsync();
            return File(streamToReadFrom, "{MIME TYPE}");

        }
        catch (Exception ex)
        {

            throw ex;
        }
        finally
        { 
        
        }
    }