How to return an Excel file from ASP.NET Core Web API web-app?

Jose Eduardo Poma Caceres picture Jose Eduardo Poma Caceres · Aug 10, 2016 · Viewed 26.8k times · Source

In similar questions, with this code works to download a PDF:

I'm testing with local files (.xlsx, .pdf, .zip) inside the Controller folder.

Similar Question Here

[HttpGet("downloadPDF")]
public FileResult TestDownloadPCF()
{
   HttpContext.Response.ContentType = "application/pdf";
   FileContentResult result = new FileContentResult
   (System.IO.File.ReadAllBytes("Controllers/test.pdf"), "application/pdf")
    {
      FileDownloadName = "test.pdf"
    };
   return result;
}

This works great!

But when another file?, for example an Excel File(.xlsx) or ZIP File(.zip), testing does not work properly.

Code :

[HttpGet("downloadOtherFile")]
public FileResult TestDownloadOtherFile()
{
  HttpContext.Response.ContentType = 
  "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet";
  FileContentResult result = new FileContentResult(System.IO.File.ReadAllBytes("Controllers/test.xlsx"), 
  "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet")
  {
    FileDownloadName = "otherfile"
   };
  return result;
}

Result: enter image description here

I also did tests with the following Content-Type:

  • "Application / vnd.ms-excel"
  • "Application / vnd.ms-excel.12"

Getting the same result.

Which is the right way to return any file type?

Thanks for your answers

Answer

BCdotWEB picture BCdotWEB · May 22, 2017

My (working) solution:

  • I've got a class that dynamically creates an XLSX file using EPPlus.Core.
    • This returns a FileInfo for the generated file's path.

This is what is in my Controller:

[HttpGet("test")]
public async Task<FileResult> Get()
{
    var contentRootPath = _hostingEnvironment.ContentRootPath;

    // "items" is a List<T> of DataObjects
    var items = await _mediator.Send(new GetExcelRequest());

    var fileInfo = new ExcelFileCreator(contentRootPath).Execute(items);
    var bytes = System.IO.File.ReadAllBytes(fileInfo.FullName);

    const string contentType = "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet";
    HttpContext.Response.ContentType = contentType;
    HttpContext.Response.Headers.Add("Access-Control-Expose-Headers", "Content-Disposition");

    var fileContentResult = new FileContentResult(bytes, contentType)
    {
        FileDownloadName = fileInfo.Name
    };

    return fileContentResult;
}

And here is what I have in Angular2:

downloadFile() {
    debugger;
    var headers = new Headers();
    headers.append('responseType', 'arraybuffer');

    let url = new URL('api/excelFile/test', environment.apiUrl);

    return this.http
        .get(url.href, {
            withCredentials: true,
            responseType: ResponseContentType.ArrayBuffer
        })
        .subscribe((response) => {
            let file = new Blob([response.blob()], { type: 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet' });
            let fileName = response.headers.get('Content-Disposition').split(';')[1].trim().split('=')[1];
            saveAs(file, fileName);
        },
        err => this.errorHandler.onError(err)
        );
}