如何在 HttpResponseMessage.Content.Headers.ContentType 中显示图像

Question

我有一个方法可以 return 图片。如下：

string fullName = dt.Rows[0]["FileName"].ToString();
                string directoryPath = MyFileMultipartFormDataStreamProvider.GetMyFileDirectory();

                var myFilePath = Path.Combine(directoryPath, fullName);
                using (FileStream fileStream = new FileStream(myFilePath, FileMode.Open))
                {
                    using (var memoryStream = new MemoryStream())
                    {
                        fileStream.CopyTo(memoryStream);

                        HttpResponseMessage result = new HttpResponseMessage(HttpStatusCode.OK);
                        result.Content = new ByteArrayContent(memoryStream.ToArray());
                        result.Content.Headers.ContentType = new MediaTypeHeaderValue("image/png");

                        return result;
                    }
                }

如何在 asp.net mvc cshtml 页面中显示图像？

Postman is ok (image).

Answer 1

我是这样解决的

byte[] mybytearray = null;
        using (var client = new HttpClient())
        {
            client.BaseAddress = new Uri(URI);

            client.DefaultRequestHeaders
                  .Accept
                  .Add(new MediaTypeWithQualityHeaderValue("image/png"));

            HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Get, "");
            try
            {
                Task<HttpResponseMessage> getResponse = client.SendAsync(request);
                HttpResponseMessage response = new HttpResponseMessage();
                response = await getResponse;
                
                if (response.IsSuccessStatusCode)
                {
                    mybytearray = response.Content.ReadAsByteArrayAsync().Result;
                }

                var responseJsonString = await response.Content.ReadAsStringAsync();
                System.Diagnostics.Debug.WriteLine(responseJsonString);
                System.Diagnostics.Debug.WriteLine("GetReportImage ReponseCode: " + response.StatusCode);
            }
            catch (Exception ex)
            {
                
            }
        }