我正在开发Windows Phone 8应用程序。我想通过PHP Web服务使用MIME类型multipart / form-data＆amp;的HTTP POST请求上传SQLite数据库。一个名为“userid = SOME_ID”的字符串数据。
我不想使用像HttpClient，RestSharp或MyToolkit这样的第三方库。我尝试了以下代码，但它没有上传文件＆amp;也没有给我任何错误。它在Android，PHP等方面运行良好，因此在Web服务中没有问题。下面是我给出的代码（适用于WP8）。怎么了？
我用谷歌搜索了，我没有具体针对WP8
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，但它也无法正常工作。
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();
}
}
}
}
答案 0 :(得分：107)
我使用MultipartFormDataContent完成了它： -
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;
答案 1 :(得分：24)
这是我最后的工作代码。我的Web服务需要一个文件（POST参数名称为“file”）＆amp;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帮助我。
答案 2 :(得分：19)
这个简单的版本也有效。
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);
}
如果需要，添加任何额外的必需标题。
答案 3 :(得分：12)
我一直在玩一点点，并提出了一个简化的，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();
};
}
你可以像这样使用它：
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，你可以处理这样的数据：
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']);
答案 4 :(得分：9)
您可以使用此课程：
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;
}
}
}
使用它：
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>';
}
答案 5 :(得分：1)
以下代码读取文件，将其转换为字节数组，然后向服务器发出请求。
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;
}
答案 6 :(得分：1)
嗨，经过一天的网上搜索，终于我用下面的源代码解决了问题 希望对您有帮助
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;
}
答案 7 :(得分：1)
在将文件作为多格式数据发送时，这对我有用：
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);
}
}
}
}
GetAuthenticatedHttpClient可以是：
private HttpClient GetAuthenticatedHttpClient()
{
HttpClient httpClient = new HttpClient();
httpClient.BaseAddress = new Uri(<yourBaseURL>));
httpClient.DefaultRequestHeaders.Add("Token, <yourToken>);
return httpClient;
}
答案 8 :(得分：0)
适用于窗口电话8.1。你可以试试这个。
Dictionary<string, object> _headerContents = new Dictionary<string, object>();
const String _lineEnd = "\r\n";
const String _twoHyphens = "--";
const String _boundary = "*****";
private async void UploadFile_OnTap(object sender, System.Windows.Input.GestureEventArgs e)
{
Uri serverUri = new Uri("http:www.myserver.com/Mp4UploadHandler", UriKind.Absolute);
string fileContentType = "multipart/form-data";
byte[] _boundarybytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _lineEnd);
byte[] _trailerbytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _twoHyphens + _lineEnd);
Dictionary<string, object> _headerContents = new Dictionary<string, object>();
SetEndHeaders(); // to add some extra parameter if you need
httpWebRequest = (HttpWebRequest)WebRequest.Create(serverUri);
httpWebRequest.ContentType = fileContentType + "; boundary=" + _boundary;
httpWebRequest.Method = "POST";
httpWebRequest.AllowWriteStreamBuffering = false; // get response after upload header part
var fileName = Path.GetFileName(MediaStorageFile.Path);
Stream fStream = (await MediaStorageFile.OpenAsync(Windows.Storage.FileAccessMode.Read)).AsStream(); //MediaStorageFile is a storage file from where you want to upload the file of your device
string fileheaderTemplate = "Content-Disposition: form-data; name=\"{0}\"" + _lineEnd + _lineEnd + "{1}" + _lineEnd;
long httpLength = 0;
foreach (var headerContent in _headerContents) // get the length of upload strem
httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes(string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value)).Length;
httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes("Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd).Length
+ Encoding.UTF8.GetBytes(_lineEnd).Length * 2 + _trailerbytes.Length;
httpWebRequest.ContentLength = httpLength + fStream.Length; // wait until you upload your total stream
httpWebRequest.BeginGetRequestStream((result) =>
{
try
{
HttpWebRequest request = (HttpWebRequest)result.AsyncState;
using (Stream stream = request.EndGetRequestStream(result))
{
foreach (var headerContent in _headerContents)
{
WriteToStream(stream, _boundarybytes);
WriteToStream(stream, string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value));
}
WriteToStream(stream, _boundarybytes);
WriteToStream(stream, "Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd);
WriteToStream(stream, _lineEnd);
int bytesRead = 0;
byte[] buffer = new byte[2048]; //upload 2K each time
while ((bytesRead = fStream.Read(buffer, 0, buffer.Length)) != 0)
{
stream.Write(buffer, 0, bytesRead);
Array.Clear(buffer, 0, 2048); // Clear the array.
}
WriteToStream(stream, _lineEnd);
WriteToStream(stream, _trailerbytes);
fStream.Close();
}
request.BeginGetResponse(a =>
{ //get response here
try
{
var response = request.EndGetResponse(a);
using (Stream streamResponse = response.GetResponseStream())
using (var memoryStream = new MemoryStream())
{
streamResponse.CopyTo(memoryStream);
responseBytes = memoryStream.ToArray(); // here I get byte response from server. you can change depends on server response
}
if (responseBytes.Length > 0 && responseBytes[0] == 1)
MessageBox.Show("Uploading Completed");
else
MessageBox.Show("Uploading failed, please try again.");
}
catch (Exception ex)
{}
}, null);
}
catch (Exception ex)
{
fStream.Close();
}
}, httpWebRequest);
}
private static void WriteToStream(Stream s, string txt)
{
byte[] bytes = Encoding.UTF8.GetBytes(txt);
s.Write(bytes, 0, bytes.Length);
}
private static void WriteToStream(Stream s, byte[] bytes)
{
s.Write(bytes, 0, bytes.Length);
}
private void SetEndHeaders()
{
_headerContents.Add("sId", LocalData.currentUser.SessionId);
_headerContents.Add("uId", LocalData.currentUser.UserIdentity);
_headerContents.Add("authServer", LocalData.currentUser.AuthServerIP);
_headerContents.Add("comPort", LocalData.currentUser.ComPort);
}
答案 9 :(得分：0)
对于在尝试以多部分形式上载时搜索403禁止问题的人们，以下内容可能会有所帮助，因为在某些情况下，由于服务器配置的不同，由于MultipartFormDataContent标头不正确，您将获得MULTIPART_STRICT_ERROR“！@eq 0”。 请注意，两个imagetag / filename变量都包含引号（\“） 例如filename =“ \” myfile.png \“”
MultipartFormDataContent form = new MultipartFormDataContent();
ByteArrayContent imageContent = new ByteArrayContent(fileBytes, 0, fileBytes.Length);
imageContent.Headers.TryAddWithoutValidation("Content-Disposition", "form-data; name="+imagetag+"; filename="+filename);
imageContent.Headers.TryAddWithoutValidation("Content-Type", "image / png");
form.Add(imageContent, imagetag, filename);
答案 10 :(得分：0)
我还想将内容上传到服务器，这是一个Spring应用程序，我终于发现我需要准确地为其设置内容类型以将其解释为文件。就是这样：
...
MultipartFormDataContent form = new MultipartFormDataContent();
var fileStream = new FileStream(uniqueTempPathInProject, FileMode.Open);
var streamContent = new StreamContent(fileStream);
streamContent.Headers.ContentType=new MediaTypeHeaderValue("application/zip");
form.Add(streamContent, "file",fileName);
...
答案 11 :(得分：0)
@loop答案的顶部。
Asp.Net MVC出现以下错误， 无法连接到远程服务器
修复： 在Web中添加以下代码后。Confing问题已为我们解决
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium import webdriver
driver=webdriver.Chrome("path of the chrome driver")
driver.get("http://udemy.com/courses/search/?src=ukw&q=python&p=1")
WebDriverWait(driver,20).until(EC.element_to_be_clickable((By.XPATH,"//span[text()='All Filters']"))).click()
WebDriverWait(driver,20).until(EC.element_to_be_clickable((By.XPATH,"//div[@class='modal-body']//label[.//span[@data-purpose='filter-option-title' and text()='Free']]/input[@type='checkbox']"))).click()
答案 12 :(得分：0)
这里是具有基本身份验证C＃的多部分数据发布
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);
}
答案 13 :(得分：0)
我知道这是老话题，但是我为此而斗争，我想分享我的解决方案。
此解决方案适用于HttpClient中的MultipartFormDataContent和System.Net.Http。您可以使用.NET Core 1.0或更高版本，或.NET Framework 4.5或更高版本来释放它。
作为一个简短的摘要，它是一个异步方法，它接收要在其中执行POST的URL，用于发送字符串的键/值集合以及用于发送文件的键/值集合作为参数。
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;
}
您可以像这样准备POST（可以根据需要添加很多字符串和文件）：
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，调用如下方法：
var result = Post(url, strings, files).GetAwaiter().GetResult();
如果需要，可以检查您的状态码，并显示以下原因：
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));
}
如果有人需要它，我在这里举一个小例子，说明如何使用PHP存储文件（在.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);
}
希望您觉得它有用，我很注意您的问题。