Send as form-data with binairy file

[LaurensVanAcker]

How can I use this library to send the data as form-data instead of application/json and add a file?
I want to implement this interface: https://developer.clearfacts.be/?shell#upload-a-sales-invoice-through-a-mutation

I have a Stream-object with the PDF-file and this:

var uploadRequest = new GraphQLRequest {
    Query = @"
        mutation upload {
            vatnumber: $vat,
            filename: $name,
            invoicetype: $invoicetype
         }",
         OperationName = "upload",
             Variables = new {
                 name = filename,
                 vat = "1234",
                 invoicetype = "SALE"
             }
         };

var graphQLUploadResponse = await graphQLClient.SendMutationAsync<File>(uploadRequest);

[rose-a]

You can propably achieve this by using a GraphQLHttpRequest instead of a GraphQLRequest and then turn the resulting HttpRequestMessage inside out with the GraphQLHttpRequest.PreprocessHttpRequestMessage action...

Interesting usecase... please let me know if you are successful!

[rose-a]

Please check out v3.0.3, the code for generating the HttpRequestMessage has been moved into GraphQLHttpRequest.

By inheriting from this class and overriding the ToHttpRequestMessage() you have full control over the HTTP request... Should be easy to create a convenient GraphQLHttpFormDataRequest now...

[LaurensVanAcker]

Thanks! It works!

internal class GraphQLHttpFormDataRequest : GraphQLHttpRequest {
   public Stream File { get; internal set; }
   public string Filename { get; internal set; }

   public override HttpRequestMessage ToHttpRequestMessage(GraphQLHttpClientOptions options, IGraphQLJsonSerializer serializer) {
   if (File is null)
       throw new ArgumentException(nameof(File));

   return new HttpRequestMessage(HttpMethod.Post, options.EndPoint) {
       Content = new MultipartFormDataContent("Upl----") {
           { new StringContent(Query), nameof(Query)},
           { new StringContent(JsonConvert.SerializeObject(Variables, NewtonsoftJsonSerializer.DefaultJsonSerializerSettings)), nameof(Variables)},
           { new StreamContent(File), Filename, Filename }
       }
   };
}