[英]Setting a “Content-Disposition” HTTP Header in Web API
我正在嘗試為上傳圖片操作創建集成測試。 從瀏覽器創建的原始請求如下所示;
POST /api/UpdateImage HTTP/1.1
Host: upload.qwe.com
Authorization: bearer KuThe6Wx/CW1TO/HVS+u3Tov3MRh8qTMDrSvQ09nMnP4OgYp
Accept-Encoding: gzip, deflate
Content-Type: multipart/form-data; boundary=Boundary-D60385FA-C164-45B0-A81E-0F6488F8E1E1
Content-Length: 375488
Accept-Language: en-us
Accept: */*
Connection: keep-alive
User-Agent: Chrome
Pragma: no-cache
Cache-Control: no-cache
--Boundary-D60385FA-C164-45B0-A81E-0F6488F8E1E1
Content-Disposition: form-data; name="fileName"
image.jpg
--Boundary-D60385FA-C164-45B0-A81E-0F6488F8E1E1
Content-Disposition: form-data; name="fileUpload"; filename="image.jpg"
Content-Type: image/jpeg
還有我的集成測試代碼;
MultipartContent multipartContent = new MultipartContent();
multipartContent.Headers.TryAddWithoutValidation("Content-Type", "multipart/form-data; boundary=Boundary-D60385FA-C164-45B0-A81E-0F6488F8E1E1");
ContentDispositionHeaderValue contentDispositionHeaderValue = new ContentDispositionHeaderValue("form-data")
{
Name = "fileName"
};
multipartContent.Headers.ContentDisposition = contentDispositionHeaderValue;
// StreamContent
FileStream fileStream = File.Open(@"./Assets/" + fileName, FileMode.Open);
StreamContent stream = new StreamContent(fileStream);
multipartContent.Add(stream);
httpRequestMessage.Content = multipartContent;
return httpRequestMessage;
但是我不能設置具有Content-Disposition: form-data; name="fileUpload"; filename="image.jpg"
的數據的第二部分Content-Disposition: form-data; name="fileUpload"; filename="image.jpg"
Content-Disposition: form-data; name="fileUpload"; filename="image.jpg"
我該如何實現?
問題摘要:
修改子HttpContent
的標頭,而不是在MultipartFormDataContent
var main = new MultipartFormDataContent(Guid.NewGuid().ToString());
HttpContent content = new StringContent("image.jpg");
content.Headers.Clear();
content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("form-data") { Name = "fileName" };
main.Add(content);
content = new StreamContent(new MemoryStream(new byte[] { 1, 2, 3 }));//your file stream, or other base64 string
content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue("image/jpeg");
content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("form-data") { Name = "fileUpload", FileName="image.jpg" };
main.Add(content);
req.Content = main;
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