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[英]How to upload file to server with HTTP POST multipart/form-data?
[英]How upload image to server with multipart form data
我想模仿這個要求
這是嗅探器的日志
-----------------------------708299735697
Content-Disposition: form-data; name="_file"
1.jpg
-----------------------------708299735697
Content-Disposition: form-data; name="file"; filename="blob"
Content-Type: image/png
‰PNG
............
那是我在csharp上的代碼。
var taimalda = DateTime.Now.Ticks;
var boundary = "------------------------" + taimalda ;
var newLine = Environment.NewLine;
var propFormat = "--" + boundary + newLine +
"Content-Disposition: form-data; name=\"{0}\"" + newLine + newLine +
"{1}" + newLine;
var fileHeaderFormat = "--" + boundary + newLine +
"Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"" +
newLine + "Content-Type: image/png";
var req = (HttpWebRequest)HttpWebRequest.Create("http://xxx.ru/new_style/flash_uploader/upload.php?fileapi"+taimalda);
req.CookieContainer = s; //
System.Net.ServicePointManager.Expect100Continue = false;
req.Referer = "http://www.xxx.ru/user/setting/set_info"; // add referer
req.UserAgent = "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:2.0) Gecko/20100101 Firefox/4.0"; // add useragent
req.Method = WebRequestMethods.Http.Post; // post request
req.ContentType = "multipart/form-data; boundary=" + boundary;
using (var reqStream = req.GetRequestStream())
{
var reqWriter = new StreamWriter(reqStream);
var tmp = string.Format(propFormat, "_file", "1.jpg");
reqWriter.Write(tmp);
tmp = string.Format(fileHeaderFormat, "file", "blob");
reqWriter.Write(tmp);
reqWriter.Flush();
}
var res = req.GetResponse();
using (var resStream = res.GetResponseStream())
{
var reader = new StreamReader(resStream);
var ext = reader.ReadToEnd();
}
但是這段代碼只發送了標頭,沒有我的文件(1.jpg)
-----------------635031060420469298 Content-Disposition: form-data; name="_file" 1.jpg --------------------------635031060420469298 Content-Disposition: form-data; name="file"; filename="blob" Content-Type: image/png
我看不到您實際讀取1.jpg文件字節並將其寫入請求流的任何地方。
您需要將文件字節寫入Stream
,而不是StreamWriter
:
using (var reqStream = req.GetRequestStream())
using (var reqWriter = new StreamWriter(reqStream))
{
reqWriter.Write(propFormat, "_file", "1.jpg");
reqWriter.Write(fileHeaderFormat, "file", "blob");
int ken = fs.Read(buffer, 0, buffer.Length);
reqStream.Write(buffer, 0, ken);
reqStream.Flush();
}
也許我的代碼會幫助您
var client = new HttpClient();
client.BaseAddress = new Uri(BASE_URL);
var multipart = new MultipartFormDataContent();
foreach(var file in files)
{
var fileContent = new ByteArrayContent(System.IO.File.ReadAllBytes(file.FullName));
multipart.Add(fileContent, "files", file.Name);
}
return client.PostAsync("Images", multipart);
遲到總比不到好
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