This seemed to work for me
var stream = new FileStream(@".\Files\test.txt", FileMode.Open,
FileAccess.Read);
var reader = new StreamReader(stream, Encoding.ASCII, true);
return new ObjectResult(reader.BaseStream);
On Tuesday, 22 August 2017 10:00:01 UTC+1, Dombat wrote:
>
> I put this on Stackoverflow yesterday, but this is probably a better place
> for it!
>
> Swagger has generated a server from an API with a method like this:
>
> [HttpGet]
> [Route("SomeRoute")]
> [SwaggerOperation("GetFile")]
> [SwaggerResponse(200, type: typeof(System.IO.Stream))]
> public virtual IActionResult GetFile()
> {
> string exampleJson = null;
> string text = "This could be the contents of a file";
>
> exampleJson = text;
> var example = exampleJson != null
> ? JsonConvert.DeserializeObject<System.IO.Stream>(exampleJson)
> : default(System.IO.Stream);
> return new ObjectResult(example);
> }
>
> If I replace string text = "This could be the contents of a file" with
> lines of code such as:
>
> var stream = new FileStream(@".\Files\test.txt", FileMode.Open,
> FileAccess.Read);string text = new
> StreamReader(stream,Encoding.ASCII,true).ReadToEnd();
>
> I get a JsonConvert exception saying "invalid characters" (because the
> string is the content of the file and not json!) If I try and return a
> stream (e.g. a fileStream) I get an exception too.
>
>
> What is the correct way to return a stream from swagger generated code in
> C#?
>
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