如何使用HTTP POST multipart/form-data将文件上传到服务器?

130
我正在开发Windows Phone 8应用程序。我想使用HTTP POST请求通过PHP web服务上传SQLite数据库,MIME类型为multipart/form-data,并使用名为"userid=SOME_ID"的字符串数据。
我不想使用第三方库,如HttpClient、RestSharp或MyToolkit。我尝试了下面的代码,但它不上传文件,也没有给我任何错误信息。它在Android、PHP等方面都很好运行,因此Web服务没有问题。以下是我提供的代码(适用于WP8)。哪里出了问题?
我已经谷歌过了,但没有找到针对WP8的具体解决方案。
async void MainPage_Loaded(object sender, RoutedEventArgs e)
{
    var file = await Windows.ApplicationModel.Package.Current.InstalledLocation.GetFileAsync(DBNAME);
    //Below line gives me file with 0 bytes, why? Should I use 
    //IsolatedStorageFile instead of StorageFile
    //var file = await ApplicationData.Current.LocalFolder.GetFileAsync(DBNAME);
    byte[] fileBytes = null;
    using (var stream = await file.OpenReadAsync())
    {
        fileBytes = new byte[stream.Size];
        using (var reader = new DataReader(stream))
        {
            await reader.LoadAsync((uint)stream.Size);
            reader.ReadBytes(fileBytes);
        }
    }

    //var res = await HttpPost(Util.UPLOAD_BACKUP, fileBytes);
    HttpPost(fileBytes);
}

private void HttpPost(byte[] file_bytes)
{
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create("http://www.myserver.com/upload.php");
    httpWebRequest.ContentType = "multipart/form-data";
    httpWebRequest.Method = "POST";
    var asyncResult = httpWebRequest.BeginGetRequestStream((ar) => { GetRequestStreamCallback(ar, file_bytes); }, httpWebRequest);  
}

private void GetRequestStreamCallback(IAsyncResult asynchronousResult, byte[] postData)  
{
    //DON'T KNOW HOW TO PASS "userid=some_user_id"  
    HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;  
    Stream postStream = request.EndGetRequestStream(asynchronousResult);  
    postStream.Write(postData, 0, postData.Length);  
    postStream.Close();  
    var asyncResult = request.BeginGetResponse(new AsyncCallback(GetResponseCallback), request);  
}  

private void GetResponseCallback(IAsyncResult asynchronousResult)  
{  
    HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;  
    HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(asynchronousResult);  
    Stream streamResponse = response.GetResponseStream();  
    StreamReader streamRead = new StreamReader(streamResponse);  
    string responseString = streamRead.ReadToEnd();  
    streamResponse.Close();  
    streamRead.Close();  
    response.Close();  
}  

我也尝试在Windows 8中解决我的问题,但它也没有起作用。
public async Task Upload(byte[] fileBytes)
{
    using (var client = new HttpClient())
    {
        using (var content = new MultipartFormDataContent("Upload----" + DateTime.Now.ToString(System.Globalization.CultureInfo.InvariantCulture)))
        {
            content.Add(new StreamContent(new MemoryStream(fileBytes)));
            //Not sure below line is true or not
            content.Add(new StringContent("userid=farhanW8"));
            using (var message = await client.PostAsync("http://www.myserver.com/upload.php", content))
            {
                var input = await message.Content.ReadAsStringAsync();
            }
        }
    }
}
16个回答

227

使用MultipartFormDataContent的基本实现:

HttpClient httpClient = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();

form.Add(new StringContent(username), "username");
form.Add(new StringContent(useremail), "email");
form.Add(new StringContent(password), "password");            
form.Add(new ByteArrayContent(file_bytes, 0, file_bytes.Length), "profile_pic", "hello1.jpg");
HttpResponseMessage response = await httpClient.PostAsync("PostUrl", form);

response.EnsureSuccessStatusCode();
httpClient.Dispose();
string sd = response.Content.ReadAsStringAsync().Result;

1
MultipartFormDataContent 仅在 WP8.1 中可用。 - LZH
21
MultipartFormDataContent在.NET 4.5中可用(不仅限于WP)。 - Merrimack
11
您可以通过流的方式添加文件,而不是将整个文件内容作为byte[]存储在内存中。 var fileStream = new FileStream(filePath, FileMode.Open); form.Add(new StreamContent(fileStream), "profile_pic"); - Bill Tarbell
1
不要调用 new HttpClient。这样会使你的软件变得不稳定。 https://www.aspnetmonsters.com/2016/08/2016-08-27-httpclientwrong/ - zolty13
1
@zolty13,回答者使用了httpClient.Dispose();。同时,建议只实例化一次HttpClient。因此,在类中实例化它一次,并在需要时使用它。在类中实例化为private static readonly HttpClient client = new HttpClient(); 参考链接-"https://learn.microsoft.com/en-us/azure/architecture/antipatterns/improper-instantiation/" - Junaid Pathan
显示剩余8条评论

34

这是我的最终工作代码。我的 Web 服务需要一个文件(POST 参数名称为 "file")和一个字符串值(POST 参数名称为 "userid")。

/// <summary>
/// Occurs when upload backup application bar button is clicked. Author : Farhan Ghumra
 /// </summary>
private async void btnUploadBackup_Click(object sender, EventArgs e)
{
    var dbFile = await ApplicationData.Current.LocalFolder.GetFileAsync(Util.DBNAME);
    var fileBytes = await GetBytesAsync(dbFile);
    var Params = new Dictionary<string, string> { { "userid", "9" } };
    UploadFilesToServer(new Uri(Util.UPLOAD_BACKUP), Params, Path.GetFileName(dbFile.Path), "application/octet-stream", fileBytes);
}

/// <summary>
/// Creates HTTP POST request & uploads database to server. Author : Farhan Ghumra
/// </summary>
private void UploadFilesToServer(Uri uri, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
{
    string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(uri);
    httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
    httpWebRequest.Method = "POST";
    httpWebRequest.BeginGetRequestStream((result) =>
    {
        try
        {
            HttpWebRequest request = (HttpWebRequest)result.AsyncState;
            using (Stream requestStream = request.EndGetRequestStream(result))
            {
                WriteMultipartForm(requestStream, boundary, data, fileName, fileContentType, fileData);
            }
            request.BeginGetResponse(a =>
            {
                try
                {
                    var response = request.EndGetResponse(a);
                    var responseStream = response.GetResponseStream();
                    using (var sr = new StreamReader(responseStream))
                    {
                        using (StreamReader streamReader = new StreamReader(response.GetResponseStream()))
                        {
                            string responseString = streamReader.ReadToEnd();
                            //responseString is depend upon your web service.
                            if (responseString == "Success")
                            {
                                MessageBox.Show("Backup stored successfully on server.");
                            }
                            else
                            {
                                MessageBox.Show("Error occurred while uploading backup on server.");
                            } 
                        }
                    }
                }
                catch (Exception)
                {

                }
            }, null);
        }
        catch (Exception)
        {

        }
    }, httpWebRequest);
}

/// <summary>
/// Writes multi part HTTP POST request. Author : Farhan Ghumra
/// </summary>
private void WriteMultipartForm(Stream s, string boundary, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
{
    /// The first boundary
    byte[] boundarybytes = Encoding.UTF8.GetBytes("--" + boundary + "\r\n");
    /// the last boundary.
    byte[] trailer = Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
    /// the form data, properly formatted
    string formdataTemplate = "Content-Dis-data; name=\"{0}\"\r\n\r\n{1}";
    /// the form-data file upload, properly formatted
    string fileheaderTemplate = "Content-Dis-data; name=\"{0}\"; filename=\"{1}\";\r\nContent-Type: {2}\r\n\r\n";

    /// Added to track if we need a CRLF or not.
    bool bNeedsCRLF = false;

    if (data != null)
    {
        foreach (string key in data.Keys)
        {
            /// if we need to drop a CRLF, do that.
            if (bNeedsCRLF)
                WriteToStream(s, "\r\n");

            /// Write the boundary.
            WriteToStream(s, boundarybytes);

            /// Write the key.
            WriteToStream(s, string.Format(formdataTemplate, key, data[key]));
            bNeedsCRLF = true;
        }
    }

    /// If we don't have keys, we don't need a crlf.
    if (bNeedsCRLF)
        WriteToStream(s, "\r\n");

    WriteToStream(s, boundarybytes);
    WriteToStream(s, string.Format(fileheaderTemplate, "file", fileName, fileContentType));
    /// Write the file data to the stream.
    WriteToStream(s, fileData);
    WriteToStream(s, trailer);
}

/// <summary>
/// Writes string to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, string txt)
{
    byte[] bytes = Encoding.UTF8.GetBytes(txt);
    s.Write(bytes, 0, bytes.Length);
}

/// <summary>
/// Writes byte array to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, byte[] bytes)
{
    s.Write(bytes, 0, bytes.Length);
}

/// <summary>
/// Returns byte array from StorageFile. Author : Farhan Ghumra
/// </summary>
private async Task<byte[]> GetBytesAsync(StorageFile file)
{
    byte[] fileBytes = null;
    using (var stream = await file.OpenReadAsync())
    {
        fileBytes = new byte[stream.Size];
        using (var reader = new DataReader(stream))
        {
            await reader.LoadAsync((uint)stream.Size);
            reader.ReadBytes(fileBytes);
        }
    }

    return fileBytes;
}

我非常感谢Darin Rousseau的帮助,他对IT技术的理解让我受益匪浅。


31

这个简化版本也可以工作。

public void UploadMultipart(byte[] file, string filename, string contentType, string url)
{
    var webClient = new WebClient();
    string boundary = "------------------------" + DateTime.Now.Ticks.ToString("x");
    webClient.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
    var fileData = webClient.Encoding.GetString(file);
    var package = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"file\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n{3}\r\n--{0}--\r\n", boundary, filename, contentType, fileData);

    var nfile = webClient.Encoding.GetBytes(package);

    byte[] resp = webClient.UploadData(url, "POST", nfile);
}

如有需要,请添加任何额外所需的标头。


如何在Windows Phone 8.1 Rt上执行它? - iamatsundere181
在与MultipartFormDataContent搏斗无果后,这种方法效果很好。虽然不太引人注目,但如果您需要对HTTP消息进行详细控制,这是一个很好的解决方案。 - John Craft
非常感谢你,伙计。我花了三天时间解决这个问题,但没有成功。这是完美的解决方案。 - user5093161
@wolf5 在上面的例子中,我想发送jpg图像,那么内容类型将是什么? - A.Goutam
查找MIME类型:https://developer.mozilla.org/zh-CN/docs/Web/HTTP/Basics_of_HTTP/MIME_types/Complete_list_of_MIME_types - Wolf5
唯一对我起作用的解决方案,非常感谢你。 - Rayyan

14

我一直在尝试着做一些实验,并想出了一种简化且更通用的解决方案:

private static string sendHttpRequest(string url, NameValueCollection values, NameValueCollection files = null)
{
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
    // The first boundary
    byte[] boundaryBytes = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "\r\n");
    // The last boundary
    byte[] trailer = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
    // The first time it itereates, we need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick
    byte[] boundaryBytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");

    // Create the request and set parameters
    HttpWebRequest request = (HttpWebRequest) WebRequest.Create(url);
    request.ContentType = "multipart/form-data; boundary=" + boundary;
    request.Method = "POST";
    request.KeepAlive = true;
    request.Credentials = System.Net.CredentialCache.DefaultCredentials;

    // Get request stream
    Stream requestStream = request.GetRequestStream();

    foreach (string key in values.Keys)
    {
        // Write item to stream
        byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}", key, values[key]));
        requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
        requestStream.Write(formItemBytes, 0, formItemBytes.Length);
    }

    if (files != null)
    { 
        foreach(string key in files.Keys)
        {
            if(File.Exists(files[key]))
            {
                int bytesRead = 0;
                byte[] buffer = new byte[2048];
                byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: application/octet-stream\r\n\r\n", key, files[key]));
                requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
                requestStream.Write(formItemBytes, 0, formItemBytes.Length);

                using (FileStream fileStream = new FileStream(files[key], FileMode.Open, FileAccess.Read))
                {
                    while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
                    {
                        // Write file content to stream, byte by byte
                        requestStream.Write(buffer, 0, bytesRead);
                    }

                    fileStream.Close();
                }
            }
        }
    }

    // Write trailer and close stream
    requestStream.Write(trailer, 0, trailer.Length);
    requestStream.Close();

    using (StreamReader reader = new StreamReader(request.GetResponse().GetResponseStream()))
    {
        return reader.ReadToEnd();
    };
}

您可以像这样使用它:
string fileLocation = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) + Path.DirectorySeparatorChar + "somefile.jpg";
NameValueCollection values = new NameValueCollection();
NameValueCollection files = new NameValueCollection();
values.Add("firstName", "Alan");
files.Add("profilePicture", fileLocation);
sendHttpRequest("http://example.com/handler.php", values, files);

在PHP脚本中,您可以像这样处理数据:
echo $_POST['firstName'];
$name = $_POST['firstName'];
$image = $_FILES['profilePicture'];
$ds = DIRECTORY_SEPARATOR;
move_uploaded_file($image['tmp_name'], realpath(dirname(__FILE__)) . $ds . "uploads" . $ds . $image['name']);

1
写得很好,将文件分块写入流中,而不是将所有内容加载到一个byte[]中并传递它。 - Bill Tarbell

11
您可以使用这个类:
using System.Collections.Specialized;
class Post_File
{
    public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
        byte[] boundarybytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");  // the first time it itereates, you need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick.  


        HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
        wr.Accept = "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8";
        var nvc2 = new NameValueCollection();
        nvc2.Add("Accepts-Language", "en-us,en;q=0.5");
        wr.Headers.Add(nvc2);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;


        Stream rs = wr.GetRequestStream();

        bool firstLoop = true;
        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            if (firstLoop)
            {
                rs.Write(boundarybytesF, 0, boundarybytesF.Length);
                firstLoop = false;
            }
            else
            {
                rs.Write(boundarybytes, 0, boundarybytes.Length);
            }
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, new FileInfo(file).Name, contentType);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

        FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
        byte[] buffer = new byte[4096];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            rs.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();

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

        WebResponse wresp = null;
        try
        {
            wresp = wr.GetResponse();
            Stream stream2 = wresp.GetResponseStream();
            StreamReader reader2 = new StreamReader(stream2);
        }
        catch (Exception ex)
        {
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }
}

使用 IT 技术:

NameValueCollection nvc = new NameValueCollection();
//nvc.Add("id", "TTR");
nvc.Add("table_name", "uploadfile");
nvc.Add("commit", "uploadfile");
Post_File.HttpUploadFile("http://example/upload_file.php", @"C:\user\yourfile.docx", "uploadfile", "application/vnd.ms-excel", nvc);

示例服务器 upload_file.php

m('File upload '.(@copy($_FILES['uploadfile']['tmp_name'],getcwd().'\\'.'/'.$_FILES['uploadfile']['name']) ? 'success' : 'failed'));
function m($msg) {
    echo '<div style="background:#f1f1f1;border:1px solid #ddd;padding:15px;font:14px;text-align:center;font-weight:bold;">';
    echo $msg;
    echo '</div>';
}

非常好,你的代码运行得很顺畅。=)这正是我寻找的适用于普通表单字段和文件字段的解决方案。你的代码也很容易扩展到多文件上传。谢谢! - tpartee
我认为如果nvc为空,你在开头发送了“太多的新段落”(“\r\n”)。 - Mike

5
以下是我在发送文件时使用mult-form数据的方法:

以下是我的成功之道:

    public T HttpPostMultiPartFileStream<T>(string requestURL, string filePath, string fileName)
    {
        string content = null;

        using (MultipartFormDataContent form = new MultipartFormDataContent())
        {
            StreamContent streamContent;
            using (var fileStream = new FileStream(filePath, FileMode.Open))
            {
                streamContent = new StreamContent(fileStream);

                streamContent.Headers.Add("Content-Type", "application/octet-stream");
                streamContent.Headers.Add("Content-Disposition", string.Format("form-data; name=\"file\"; filename=\"{0}\"", fileName));
                form.Add(streamContent, "file", fileName);

                using (HttpClient client = GetAuthenticatedHttpClient())
                {
                    HttpResponseMessage response = client.PostAsync(requestURL, form).GetAwaiter().GetResult();
                    content = response.Content.ReadAsStringAsync().GetAwaiter().GetResult();



                    try
                    {
                        return JsonConvert.DeserializeObject<T>(content);
                    }
                    catch (Exception ex)
                    {
                        // Log the exception
                    }

                    return default(T);
                }
            }
        }
    }

上述使用的GetAuthenticatedHttpClient可以:

private HttpClient GetAuthenticatedHttpClient()
{
    HttpClient httpClient = new HttpClient();
    httpClient.BaseAddress = new Uri(<yourBaseURL>));
    httpClient.DefaultRequestHeaders.Add("Token, <yourToken>);
    return httpClient;
}

1
谢谢您提供 MultipartFormDataContent 的想法,这正是我所需要的。 - Nasreddine Galfout

3

我知道这是一个旧的线程,但我一直在解决这个问题,我想分享我的解决方案。

这个解决方案使用了System.Net.Http中的HttpClientMultipartFormDataContent。您可以将其用于.NET Core 1.0或更高版本,或.NET Framework 4.5或更高版本。

简单来说,这是一个异步方法,它接收三个参数:您要执行POST请求的URL、发送字符串的键值对集合和发送文件的键值对集合。

private static async Task<HttpResponseMessage> Post(string url, NameValueCollection strings, NameValueCollection files)
{
    var formContent = new MultipartFormDataContent(/* If you need a boundary, you can define it here */);

    // Strings
    foreach (string key in strings.Keys)
    {
        string inputName = key;
        string content = strings[key];

        formContent.Add(new StringContent(content), inputName);
    }

    // Files
    foreach (string key in files.Keys)
    {
        string inputName = key;
        string fullPathToFile = files[key];

        FileStream fileStream = File.OpenRead(fullPathToFile);
        var streamContent = new StreamContent(fileStream);
        var fileContent = new ByteArrayContent(streamContent.ReadAsByteArrayAsync().Result);
        formContent.Add(fileContent, inputName, Path.GetFileName(fullPathToFile));
    }

    var myHttpClient = new HttpClient();
    var response = await myHttpClient.PostAsync(url, formContent);
    //string stringContent = await response.Content.ReadAsStringAsync(); // If you need to read the content

    return response;
}

你可以像这样准备你的POST请求(你可以添加任意数量的字符串和文件):

string url = @"http://yoursite.com/upload.php"

NameValueCollection strings = new NameValueCollection();
strings.Add("stringInputName1", "The content for input 1");
strings.Add("stringInputNameN", "The content for input N");

NameValueCollection files = new NameValueCollection();
files.Add("fileInputName1", @"FullPathToFile1"); // Path + filename
files.Add("fileInputNameN", @"FullPathToFileN");

最后,按照以下方式调用该方法:

var result = Post(url, strings, files).GetAwaiter().GetResult();

如果您想要检查状态码并显示原因,可以按如下方式进行:
if (result.StatusCode == HttpStatusCode.OK)
{
    // Logic if all was OK
}
else
{
    // You can show a message like this:
    Console.WriteLine(string.Format("Error. StatusCode: {0} | ReasonPhrase: {1}", result.StatusCode, result.ReasonPhrase));
}

如果有人需要,这里提供一个PHP接收并存储文件的小例子(在我们.NET应用程序的另一侧):

<?php

if (isset($_FILES['fileInputName1']) && $_FILES['fileInputName1']['error'] === UPLOAD_ERR_OK)
{
  $fileTmpPath = $_FILES['fileInputName1']['tmp_name'];
  $fileName = $_FILES['fileInputName1']['name'];

  move_uploaded_file($fileTmpPath, '/the/final/path/you/want/' . $fileName);
}

希望你能发现它有用,我会注意你的问题。


Async().Result 不是异步的。 - OwnageIsMagic
@OwnageIsMagic 我无法理解你指的是代码的哪一部分。让我们来看看这里发生事情的顺序:1)在 HttpClient() 对象中使用 PostAsync(...),它是异步的。2)由于 1),我的 Post 方法返回一个异步任务。3)由于 2),我使用 GetAwaiter().GetResult() 将我的 Post(...) 方法与钩子连接起来,等待异步任务结束。我错过了什么吗?干杯 - cesAR
我指的是 streamContent.ReadAsByteArrayAsync().Result。 但是 GetAwaiter().GetResult() 也不是异步的。 - OwnageIsMagic

2

大家好,经过一天的网上搜索,我终于用下面的源代码解决了问题。希望能对你有所帮助。

Original Answer翻译成"最初的回答"

    public UploadResult UploadFile(string  fileAddress)
    {
        HttpClient client = new HttpClient();

        MultipartFormDataContent form = new MultipartFormDataContent();
        HttpContent content = new StringContent("fileToUpload");
        form.Add(content, "fileToUpload");       
        var stream = new FileStream(fileAddress, FileMode.Open);            
        content = new StreamContent(stream);
        var fileName = 
        content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
        {
            Name = "name",
            FileName = Path.GetFileName(fileAddress),                 
        };
        form.Add(content);
        HttpResponseMessage response = null;          

        var url = new Uri("http://192.168.10.236:2000/api/Upload2");
        response = (client.PostAsync(url, form)).Result;          

    }

2

这里是使用基本身份验证的C#多部分数据post请求

public string UploadFilesToRemoteUrl(string url)
    {
        try
        {                             

            Dictionary<string, object> formFields = new Dictionary<string, object>();
            formFields.Add("requestid", "{\"id\":\"idvalue\"}");

            string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");

            HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
            request.ContentType = "multipart/form-data; boundary=" + boundary;

            // basic authentication.
            var username = "userid";
            var password = "password";

            string credidentials = username + ":" + password;
            var authorization = Convert.ToBase64String(Encoding.Default.GetBytes(credidentials));
            request.Headers["Authorization"] = "Basic " + authorization;

            request.Method = "POST";
            request.KeepAlive = true;

            Stream memStream = new System.IO.MemoryStream();
            WriteFormData(formFields, memStream, boundary);

            FileInfo fileToUpload = new FileInfo(@"filelocation with name");
            string fileFormKey = "file";
            if (fileToUpload != null)
            {
                WritefileToUpload(fileToUpload, memStream, boundary, fileFormKey);
            }
            request.ContentLength = memStream.Length;

            using (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);
            }

            using (var response = request.GetResponse())
            {
                Stream responseSReam = response.GetResponseStream();
                StreamReader streamReader = new StreamReader(responseSReam);
                return streamReader.ReadToEnd();
            }
        }
        catch (WebException ex)
        {
            using (WebResponse response = ex.Response)
            {
                HttpWebResponse httpResponse = (HttpWebResponse)response;
                using (var streamReader = new StreamReader(response.GetResponseStream()))
                    return streamReader.ReadToEnd();

            }
        }
    }

    // write form id.
    public static void WriteFormData(Dictionary<string, object> dictionary, Stream stream, string mimeBoundary)
    {
        string formdataTemplate = "\r\n--" + mimeBoundary +
                                    "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
        if (dictionary != null)
        {
            foreach (string key in dictionary.Keys)
            {
                string formitem = string.Format(formdataTemplate, key, dictionary[key]);
                byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
                stream.Write(formitembytes, 0, formitembytes.Length);
            }
        }
    }

    // write file.
    public static void WritefileToUpload(FileInfo file, Stream stream, string mimeBoundary, string formkey)
    {
        var boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + mimeBoundary + "\r\n");
        var endBoundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + mimeBoundary + "--");

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n" +
                                "Content-Type: application/octet-stream\r\n\r\n";

        stream.Write(boundarybytes, 0, boundarybytes.Length);
        var header = string.Format(headerTemplate, formkey, file.Name);
        var headerbytes = System.Text.Encoding.UTF8.GetBytes(header);

        stream.Write(headerbytes, 0, headerbytes.Length);

        using (var fileStream = new FileStream(file.FullName, FileMode.Open, FileAccess.Read))
        {
            var buffer = new byte[1024];
            var bytesRead = 0;
            while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
            {
                stream.Write(buffer, 0, bytesRead);
            }
        }
        stream.Write(endBoundaryBytes, 0, endBoundaryBytes.Length);
    } 

2
下面的代码读取一个文件,将其转换为字节数组,然后发送请求到服务器。
    public void PostImage()
    {
        HttpClient httpClient = new HttpClient();
        MultipartFormDataContent form = new MultipartFormDataContent();

        byte[] imagebytearraystring = ImageFileToByteArray(@"C:\Users\Downloads\icon.png");
        form.Add(new ByteArrayContent(imagebytearraystring, 0, imagebytearraystring.Count()), "profile_pic", "hello1.jpg");
        HttpResponseMessage response = httpClient.PostAsync("your url", form).Result;

        httpClient.Dispose();
        string sd = response.Content.ReadAsStringAsync().Result;
    }

    private byte[] ImageFileToByteArray(string fullFilePath)
    {
        FileStream fs = File.OpenRead(fullFilePath);
        byte[] bytes = new byte[fs.Length];
        fs.Read(bytes, 0, Convert.ToInt32(fs.Length));
        fs.Close();
        return bytes;
    }

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