Xml vb.net中的REST调用：如何传递键；文件"；这在卷曲中起作用

我已经为这个问题挣扎了将近一个月了，我在网上看到了所有我能找到的东西，但没有解决办法。我的问题是：我正在为RESTful API服务实现一个客户端，该服务必须在vb.net中通过POST调用发送XML文件。我可以让它在获取xml格式的数据时正常工作，但在发送这个xml文件时，我总是会遇到“400错误请求错误”

我已经弄明白，这必须是一个必须传递到服务器的密钥的问题（显然，服务器只接受文件上传以供发布，我不能将其作为字符串发送）

基本上，这个调用适用于cURL，但我正在努力在vb.net中实现我自己的调用，传递正确的值

工作cURL调用：（成功传输XML）

无法运行Vb.net代码：（这给了我400个错误请求）

在任何情况下，我也尝试了其他2个客户端（POSTMAN和REST控制台，Google Chrome的2个扩展），只有在“key”字段中添加值“file”，我才能让它工作。我必须插入特定的4个字符“文件”才能使其正常工作。所以，问题是：如何在Vb.net调用中添加相同的值？如何在工作的Vb.net代码中翻译cURL调用的代码？非常感谢您的时间和帮助

找到我想在这里添加的东西的图像

另外，我不能使用PUT，我必须使用POST（服务器限制）

我还添加了适用于我的目的的HTML代码，从我的pc使用服务器（再次查看“文件”键）

---

我认为您的问题在于您不知道请求应该是什么样子，请使用fiddler查看cURL发送的请求，并尝试使用vb.net实现相同的请求。在我看来，您的secons代码（带有缓冲区，而不是xml序列化程序）应该可以工作。。。但只有在服务器端可以理解这种结构的情况下

您应该尝试将文件作为原始数据，作为二进制缓冲区发送。不要使用.net序列化程序或类似的东西。我不明白“file”键字段是什么意思，这是属于HTTP头还是HTTP消息体中发送的数据对象中的值？->在这两种情况下，使用vb.net或c#Fiddler将它们添加到消息中应该不难。Fiddler让我很高兴。谢谢拉法！

  c:>curl -u username:password -F "file=@filename.xml" -X POST http://hostname.com/URI?parameters

Dim ss As String = "" 'server says...
Dim S As String = txb_username.Text & ":" & txb_password.Text
Dim EncodedString As String = System.Convert.ToBase64String(System.Text.Encoding.UTF8.GetBytes(S))
Dim req As HttpWebRequest = Nothing
Dim res As HttpWebResponse = Nothing

Try
    Dim xmlDoc As System.Xml.XmlDocument = New System.Xml.XmlDocument
    xmlDoc.XmlResolver = Nothing

    xmlDoc.Load("c:\path\file4.xml")

    Dim sXML As String = "file" & xmlDoc.InnerXml '<- This is where I try to put the "KEY"

    Dim url As String = "http:/host.com+URI"

    req = CType(WebRequest.Create(url), Net.HttpWebRequest) 'or Directcast ... 
    req.Method = "POST"
    req.Headers.Add("Authorization: Basic " & EncodedString)
    req.ContentType = "multipart/form-data"
    req.ContentLength = sXML.Length 
    req.Accept = "*/*"

    System.Windows.Forms.Application.DoEvents()

    Dim sw As System.IO.StreamWriter = New System.IO.StreamWriter(req.GetRequestStream)
    StatusUpdate(sXML)
    sw.Write(sXML)
    sw.Close()
    ss = "server says: "
    res = CType(req.GetResponse, HttpWebResponse)
     StatusUpdate(req.ToString)
    Catch ex As Exception
        StatusUpdate(ss & ex.Message)
    Finally

End Try

    Dim requestStream As Stream = Nothing
    Dim fileStream As FileStream = Nothing
    Dim uploadResponse As Net.HttpWebResponse = Nothing

    Try

        Dim uploadRequest As Net.HttpWebRequest = CType(Net.HttpWebRequest.Create(URI.Text & Uri_part2.text), Net.HttpWebRequest)

        uploadRequest.Method = Net.WebRequestMethods.Http.Post
        uploadRequest.ContentType = "text/xml; charset=utf-8"
        uploadRequest.Credentials = New NetworkCredential("user", "pass")
        uploadRequest.KeepAlive = True
        uploadRequest.UserAgent = "User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.0.10) Gecko/2009042316 Firefox/3.0.10"
        uploadRequest.Accept = ("text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8")
        uploadRequest.Headers.Add("Accept-Language: en-us,en;q=0.5")
        uploadRequest.Headers.Add("Accept-Encoding: gzip,deflate")
        uploadRequest.Headers.Add("Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7")
        uploadRequest.Headers.Add("Content-Disposition: form-data; name=""file"";")
        uploadRequest.ContentType = "application/xml; charset=utf-8"

        requestStream = uploadRequest.GetRequestStream()
        fileStream = File.Open("C:\example.xml", FileMode.Open)

        Dim a As Integer

        Dim buffer(1024) As Byte
        Dim bytesRead As Integer
        While True
            a = a + 1
            bytesRead = fileStream.Read(buffer, 0, buffer.Length)
            StatusUpdate(buffer(a))

            If bytesRead = 0 Then
                Exit While
            End If
            requestStream.Write(buffer, 0, bytesRead)

        End While

       requestStream.Close()

        uploadResponse = uploadRequest.GetResponse()
        Dim responseReader As StreamReader = New StreamReader(uploadRequest.GetResponse.GetResponseStream())
        Dim x As String = responseReader.ReadToEnd()
        responseReader.Close()
        StatusUpdate(x)


    Catch ex As UriFormatException
        StatusUpdate("UriFormatException: " & ex.Message)
    Catch ex As IOException
        StatusUpdate("IOException: " & ex.Message)
    Catch ex As Net.WebException
        StatusUpdate("Net.WebException: " & ex.Message)

    Finally

        If uploadResponse IsNot Nothing Then
            uploadResponse.Close()
        End If

        If fileStream IsNot Nothing Then
            fileStream.Close()
        End If

        If requestStream IsNot Nothing Then
            requestStream.Close()
        End If

    End Try

<html>
<body>
<form enctype="multipart/form-data" action="http://URI" method="POST">
<table border=0>

<tr>
<td align="right">File&nbsp;</td>
<td><input type="FILE" name="file"></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input type="submit"></td>
</tr>
</table>
</form>
</body>
</html>

  #!perl

use strict;
use LWP; # Loads all important LWP classes

my $client_id         = 1234;
my $filename          = "new_file.xml";

### Prepare to make a request

my $browser = LWP::UserAgent->new;

my $url = "http://uri.com?&xx=$client_id";

my @post_pairs = (
    #'client_id_in' => $client_id,
    'file' => [$filename],
);

my @ns_headers = (
    'User-Agent' => 'Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.0.10) Gecko/2009042316 Firefox/3.0.10',
    'Accept' => 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
    'Accept-Language' => 'en-us,en;q=0.5',
    'Accept-Encoding' => 'gzip,deflate',
    'Accept-Charset' => 'ISO-8859-1,utf-8;q=0.7,*;q=0.7',
    'Authorization' => 'Basic base64EncodedCredentialsHere',
    'Content_Type' => 'form-data',
);

### Make a request
my $response = $browser->post($url, \@post_pairs, @ns_headers);
die "Can't get $url -- ", $response->status_line
        unless $response->is_success;

### Display the response
print STDOUT $response->content;