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我正在尝试使用 api 上传图像配置文件，但出现 Unknown: NOT_FOUND 404 错误。我正在使用的调用是 POST /d2l/api/lp/1.3/profile/(profileId)/image，我正在传递内容类型、长度和文件名 (profileImage)。我将图像作为数据流传递。我也缩小了图像的大小。有任何想法吗？

                public static void UploadFilesToRemoteUrl(string file, string logpath, NameValueCollection nvc, ID2LUserContext userContext, string accion)
    {
        var uri = userContext.CreateAuthenticatedUri(accion, "POST");
        string boundary = "bde472ff1f1a46539e54e655857c27c1";

        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(uri);
        request.ContentType = "multipart/form-data; boundary=" +
        boundary;
        request.Headers.Add("Accept-Encoding", "gzip, deflate, compress");
        request.Method = "POST";
        request.KeepAlive = true;

        request.Proxy.Credentials = new NetworkCredential(Constantes.UsuarioProxy, Constantes.PasswordProxy, Constantes.DominioProxy);

        Stream memStream = new System.IO.MemoryStream();

        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" +
        boundary + "\r\n");


        string formdataTemplate = "\r\n--" + boundary +
        "\r\nContent-Disposition: form-data; name=\"profileImage\"; filename=\"profileImage.png\" \r\nContent-Type: image/png\r\n";

        byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formdataTemplate);
        memStream.Write(formitembytes, 0, formitembytes.Length);


        // Read image File *************************************************************
        FileStream fileStream = new FileStream(file, FileMode.Open,FileAccess.Read);
        byte[] buffer = new byte[1024];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            memStream.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();
        //*****************************************************************************
        //*********** End Read image file *********************************************

        memStream.Write(boundarybytes, 0, boundarybytes.Length);

        request.ContentLength = memStream.Length;
        Stream requestStream = request.GetRequestStream();
        memStream.Position = 0;
        byte[] tempBuffer = new byte[memStream.Length];
        memStream.Read(tempBuffer, 0, tempBuffer.Length);
        memStream.Close();
        requestStream.Write(tempBuffer, 0, tempBuffer.Length);
        requestStream.Close();

        HttpWebResponse response = (HttpWebResponse)request.GetResponse();
        StreamReader reader = new StreamReader(response.GetResponseStream(), Encoding.UTF8);
        string responseValence = reader.ReadToEnd();

    }
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2 回答 2

0

固定的。此代码正常工作：

        public static void UploadFilesToRemoteUrl(byte[] profileImage, ID2LUserContext userContext, string accion)
    {
        //Reference:
        //action = "/d2l/api/lp/1.3/profile/" + profileIdentifier + "/image";

        //profileImage = the profileImage of user read from disk:
        /*
        FileStream fileStream = new FileStream(pictureLocalPath, FileMode.Open, FileAccess.Read);
        Byte[] img = new Byte[fileStream.Length];
        fileStream.Read(img, 0, Convert.ToInt32(img.Length));
        fileStream.Close();
        */

        var uri = userContext.CreateAuthenticatedUri(accion, "POST");
        string boundary = "bde472ff1f1a46539e54e655857c27c1";

        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(uri);
        request.ContentType = "multipart/form-data; boundary=" +
        boundary;
        request.Headers.Add("Accept-Encoding", "gzip, deflate, compress");
        request.Method = "POST";
        request.KeepAlive = true;

        request.Proxy.Credentials = new NetworkCredential(Constantes.UsuarioProxy, Constantes.PasswordProxy, Constantes.DominioProxy);

        Stream memStream = new System.IO.MemoryStream();

        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" +
        boundary + "\r\n");


        string formdataTemplate = "\r\n--" + boundary +
        "\r\nContent-Disposition: form-data; name=\"profileImage\"; filename=\"profileImage.jpg\"\r\nContent-Type: image/jpeg;\r\n\r\n";

        byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formdataTemplate);
        memStream.Write(formitembytes, 0, formitembytes.Length);

        //escribo el array de byte de la imagen
        memStream.Write(profileImage, 0, profileImage.Length);

        byte[] boundaryClose = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--");
        memStream.Write(boundaryClose, 0, boundarybytes.Length);

        StreamReader readerReq = new StreamReader(memStream);
        string stringReq = readerReq.ReadToEnd();

        request.ContentLength = memStream.Length;
        Stream requestStream = request.GetRequestStream();
        memStream.Position = 0;
        byte[] tempBuffer = new byte[memStream.Length];
        memStream.Read(tempBuffer, 0, tempBuffer.Length);
        memStream.Close();
        requestStream.Write(tempBuffer, 0, tempBuffer.Length);
        requestStream.Close();

        HttpWebResponse response = (HttpWebResponse)request.GetResponse();
        if (response.StatusCode == HttpStatusCode.OK)
        {
            StreamReader reader = new StreamReader(response.GetResponseStream(), Encoding.UTF8);
            string responseValence = reader.ReadToEnd();
            response.Close();
        }
    }
于 2014-02-03T18:32:47.603 回答
0

404 似乎最有可能来自

  • 您提供的 API 路由（并且后端服务无法将您的 API 路由与处理程序方法匹配）：这可能是由于profileId值不正确、路由输入错误或路由中的 API 版本号不正确，等等。

  • 出于某种原因，后端服务正在接受您的个人资料图像数据，但无法将其分配给用户的个人资料。

这是上传的个人资料图像数据包的一些请求/响应详细信息的捕获。将个人资料图片上传到“我的个人资料”时，我使用从这些值构建的 HTTP 标头：

{'Content-Length': '75143', 
 'User-Agent': 'python-requests/2.2.1 CPython/3.3.3 Darwin/12.5.0', 
 'Content-Type': 'multipart/form-data; boundary=bde472ff1f1a46539e54e655857c27c1', 
 'Accept': '*/*', 
 'Accept-Encoding': 'gzip, deflate, compress'}

请注意，这是一个多部分/表单数据内容主体，在单个主体部分周围有一个边界标记。请求的正文内容如下所示：

--bde472ff1f1a46539e54e655857c27c1
Content-Disposition: form-data; name="profileImage"; filename="profile_img-225x225.png"

[actual PNG bytes here]
--bde472ff1f1a46539e54e655857c27c1--

Content-Disposition 标头中的属性name必须是profileImage，并且该filename属性应以您用于提供内容的本地文件名命名（因此，就后端服务而言，它的值并不是特别相关） .

最后，有特定的角色权限允许用户编辑其他人的个人资料图像，因此您应该确保 API 调用的调用用户上下文具有编辑其他人图像的权限。

于 2014-01-31T21:14:49.023 回答