CODEWITHSUNDEEP
asp.net-web-apiodata
Navap
Navap

Reputation: 1072

Media Resource Support for OData POST in Web API

I would like to create oData controller to upload files

FileDto

  1. FileId
  2. NameWithExtension (Type: String)
  3. Metadata (Type: List)
  4. Content (Type: Stream)

=========================Http Request Actions==================

• GET: ~/Files({id})

Content-Type: application/json
Result: FileDto without Content

• GET: ~/Files({id})

Content-Type: application/octet-stream
Result: Stream of the File only

• POST: ~/Files

Content-Type: ?
Body: FileDto with Content
Result: FileId

Not sure how i can achieve this when coupled with OData.

Thanks in advance

Upvotes: 13

Views: 1353

Answers (1)

Ricardo Valente
Ricardo Valente

Reputation: 591

This page explains how to create an oDataController.

1) To install the package on your project, open the console manager and type this: 

Install-Package Microsoft.AspNet.Odata

2) Open your WebApiConfig.cs and, inside Register method, add this code:

 ODataModelBuilder builder = new ODataConventionModelBuilder();
            builder.EntitySet<FileDto>("File");            
            config.MapODataServiceRoute(
                routeName: "ODataRoute",
                routePrefix: null,
                model: builder.GetEdmModel());

3) Create your oDataController replacing the yourDataSourceHere to use your own class:

public class FileController : ODataController
{
    [EnableQuery]
    public IQueryable<FileDto> Get()
    {
        return yourDataSourceHere.Get();
    }

    [EnableQuery]
    public SingleResult<FileDto> Get([FromODataUri] int key)
    {
        IQueryable<FileDto> result = yourDataSourceHere.Get().Where(p => p.Id == key);
        return SingleResult.Create(result);
    }

    public IHttpActionResult Post(FileDto File)
    {
        if (!ModelState.IsValid)
            return BadRequest(ModelState);

        yourDataSourceHere.Add(product);

        return Created(File);
    }
}

OBS: To test this solution, I changed the FileDto's property Content. More specifically, it's type! From Stream to byte[]. Posted the content as Base64 string.

Upvotes: 3

Related Questions