I have included a link on my website to download images. When I click on the link I would like the download to automatically start.
Currently when I click on the link I’m getting back the response message: Example:
StatusCode: 200, ReasonPhrase: 'OK', Version: 1.1, Content: System.Net.Http.PushStreamContent, Headers: { Content-Type: application/octet-stream Content-Disposition: attachment; filename=895621d7-57a4-47a5-8dc5-ae36a2623826Banneraaaaaaaa.jpg }
How do I modify the code below to start the download automatically. I think I might be returning the wrong type:
Here is my code:
public HttpResponseMessage DownloadImageFile(string filepath)
{
filepath = "https://mysite.com/" + filepath;
try
{
var response = new HttpResponseMessage();
response.Content = new PushStreamContent((Stream outputStream, HttpContent content, TransportContext context) =>
{
try
{
DownloadFile(filepath, outputStream);
}
finally
{
outputStream.Close();
}
});
response.Content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue("application/octet-stream");
response.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
response.Content.Headers.ContentDisposition.FileName = Path.GetFileName(filepath);
return response;
}
catch (Exception ex)
{
}
return null;
}
public void DownloadFile(string file, Stream response)
{
var bufferSize = 1024 * 1024;
using (var stream = new FileStream(file, FileMode.Open, FileAccess.Read))
{
var buffer = new byte[bufferSize];
var bytesRead = 0;
while ((bytesRead = stream.Read(buffer, 0, bufferSize)) > 0)
{
response.Write(buffer, 0, bytesRead);
}
response.Flush();
}
}
}
You should use one of the Controller.File overload. The File() helper method provides support for returning the contents of a file. The MediaTypeNames class can be used to get the MIME type for a specific file name extension.
For example:
public FileResult Download(string fileNameWithPath)
{
// Option 1 - Native support for file read
return File(fileNameWithPath, System.Net.Mime.MediaTypeNames.Application.Octet, Path.GetFileName(fileNameWithPath));
// Option 2 - Read byte array and pass to file object
//byte[] fileBytes = System.IO.File.ReadAllBytes(fileName); return
//File(fileBytes, System.Net.Mime.MediaTypeNames.Application.Octet,
//fileName);
}
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