C＃HttpClient 4.5 multipart / form-データのアップロード

C Httpclient 4 5 Multipart Form Data Upload

元のリンク： https://stackoverflow.com/questions/16416601/c-sharp-httpclient-4-5-multipart-form-data-upload
https://blog.csdn.net/xcymorningsun/article/details/52950107

private static void Upload() { using (var client = new HttpClient()) { client.DefaultRequestHeaders.Add('User-Agent', 'CBS Brightcove API Service') using (var content = new MultipartFormDataContent()) { var path = @'C:B2BAssetRootfiles596086596086.1.mp4' string assetName = Path.GetFileName(path) var request = new HTTPBrightCoveRequest() { Method = 'create_video', Parameters = new Params() { CreateMultipleRenditions = 'true', EncodeTo = EncodeTo.Mp4.ToString().ToUpper(), Token = 'x8sLalfXacgn-4CzhTBm7uaCxVAPjvKqTf1oXpwLVYYoCkejZUsYtg..', Video = new Video() { Name = assetName, ReferenceId = Guid.NewGuid().ToString(), ShortDescription = assetName } } } //Content-Disposition: form-data name='json' var stringContent = new StringContent(JsonConvert.SerializeObject(request)) stringContent.Headers.Add('Content-Disposition', 'form-data name='json'') content.Add(stringContent, 'json') FileStream fs = File.OpenRead(path) var streamContent = new StreamContent(fs) streamContent.Headers.Add('Content-Type', 'application/octet-stream') //Content-Disposition: form-data name='file' filename='C:B2BAssetRootfiles596090596090.1.mp4' streamContent.Headers.Add('Content-Disposition', 'form-data name='file' filename='' + Path.GetFileName(path) + ''') content.Add(streamContent, 'file', Path.GetFileName(path)) //content.Headers.ContentDisposition = new ContentDispositionHeaderValue('attachment') Task message = client.PostAsync('http://api.brightcove.com/services/post', content) var input = message.Result.Content.ReadAsStringAsync() Console.WriteLine(input.Result) Console.Read() } } }

総括する

[クライアント]サーバーにパラメーターを渡す方法は？

//Content-Disposition: form-data name='json' var stringContent = new StringContent(JsonConvert.SerializeObject(request)) stringContent.Headers.Add('Content-Disposition', 'form-data name='json'') content.Add(stringContent, 'json')

【サーバー】クライアントのパラメータを受け取る方法は？

string id=HttpContext.Current.Request['id'] string name = HttpContext.Current.Request['name']

【サーバー】転送用ファイルの受け取り方法は？

HttpContent bytesContent = new ByteArrayContent(paramFileBytes) using (var client = new HttpClient()) using (var formData = new MultipartFormDataContent()) { bytesContent.Headers.ContentDisposition = new ContentDispositionHeaderValue('form-data') { Name = 'file', // form value name must be 'file' FileName = fileName, } var response = client.PostAsync(url, formData).Result if (!response.IsSuccessStatusCode) { return null } return response.Content.ReadAsStreamAsync().Result }