本片主要解決“type=file時是可行的,type=image時不知為何總是提示【系統繁忙】”的問題。
釘釘上傳素材的API文檔寫的比較簡單,尤其是http頭那塊完全沒寫,demo也沒有.net的,杯具啊。
看了一位仁兄的帖子,原理寫的很清楚,大家可以參考一下。
http://blog.csdn.net/xxdddail/article/details/51983487
但有一個問題沒有解決,文章里也提到了,就是type=file時是可行的,type=image時不知為何總是提示【系統繁忙】
研究了好幾天,參考了下面這位仁兄的代碼
https://github.com/icodegame/DTSigninRemind
終於找到了問題,貌似是C#提供的WebClient的問題,改用HttpWebRequest解決了。
在自定義頭一樣的情況下WebClient不行,但HttpWebRequest可以。
想了解http頭的話,請看這一篇,http://blog.csdn.net/xiaojianpitt/article/details/6856536
上面github上的仁兄代碼是異步的,我需要同步方式,就直接改成同步的了。如果需要異步的代碼可以用上面鏈接里的代碼。
代碼如下
1 public static string PostFile(string url, string fileWithPath) 2 { 3 var result = string.Empty; 4 var request = (HttpWebRequest)WebRequest.Create(url); 5 var boundary = "----------" + DateTime.Now.Ticks.ToString("x"); 6 request.ContentType = "multipart/form-data; boundary=" + boundary; 7 request.Method = "POST"; 8 using (Stream requestStream = request.GetRequestStream()) 9 { 10 byte[] boundarybytes = Encoding.UTF8.GetBytes("--" + boundary + "\r\n"); 11 byte[] trailer = Encoding.UTF8.GetBytes("\r\n--" + boundary + "–-\r\n"); 12 var filename = Path.GetFileName(fileWithPath); 13 using (FileStream fs = new FileStream(fileWithPath, FileMode.Open, FileAccess.Read)) 14 { 15 byte[] bArr = new byte[fs.Length]; 16 fs.Read(bArr, 0, bArr.Length); 17 requestStream.Write(boundarybytes, 0, boundarybytes.Length); 18 var header = $"Content-Disposition:form-data;name=\"media\";filename=\"{filename}\"\r\nfilelength=\"{fs.Length}\"\r\nContent-Type:application/octet-stream\r\n\r\n"; 19 byte[] postHeaderBytes = Encoding.UTF8.GetBytes(header.ToString()); 20 requestStream.Write(postHeaderBytes, 0, postHeaderBytes.Length); 21 fs.Close(); 22 requestStream.Write(bArr, 0, bArr.Length); 23 requestStream.Write(trailer, 0, trailer.Length); 24 } 25 } 26 var response = (HttpWebResponse)request.GetResponse(); 27 var responseStream = response.GetResponseStream(); 28 using (var streamReader = new StreamReader(response.GetResponseStream())) 29 { 30 result = streamReader.ReadToEnd(); 31 } 32 return result; 33 }
調用代碼
1 var fileName = @"d:\a.jpg"; 2 var url = $"https://oapi.dingtalk.com/media/upload?access_token={access_token}&type=image"; 3 var requestResult = PostFile(url, fileName);
返回的結果