Question

我正在尝试将.zip文件从我的VB表单上传到服务器，但文件在此过程中丢失了标题信息，我无法找到错误。调试，我发现丢失的数据来自标头。有谁知道为什么会这样？

我的VB.NET代码是：

Imports System.Text

Public Class Form1

    Dim hola As String = "hola"

    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click

        HttpUploadFile("http://localhost/sitio_victor/files.php?parameter1=""value1""&parameter2=" & hola & "", "C:\MDPScan\ParaEnviar\C001_C002_C005_01062016_092619_76CDB2BAD9582D23C1F6F4D868218D6C.zip", "file", "multipart/form-data")

    End Sub

    Private Sub HttpUploadFile( _
    ByVal uri As String, _
    ByVal filePath As String, _
    ByVal fileParameterName As String, _
    ByVal contentType As String)

        Dim process As New Process
        Dim boundary As String = "---------------------------" & DateTime.Now.Ticks.ToString("x")
        Dim newLine As String = System.Environment.NewLine
        Dim boundaryBytes As Byte() = Encoding.ASCII.GetBytes(newLine & "--" & boundary & newLine)
        Dim request As Net.HttpWebRequest = Net.WebRequest.Create(uri)
        request.ContentType = "multipart/form-data; boundary=" & boundary
        request.Method = "POST"
        request.KeepAlive = True
        request.Credentials = Net.CredentialCache.DefaultCredentials
        Using requestStream As IO.Stream = request.GetRequestStream()
            Dim formDataTemplate As String = "Content-Disposition: form-data; name=""{0}""{1}{1}{2}"
            requestStream.Write(boundaryBytes, 0, boundaryBytes.Length)


            Dim headerTemplate As String = "Content-Disposition: form-data; name=""{0}""; filename=""{1}""{2}Content-Type: {3};"
            Dim header As String = String.Format(headerTemplate, fileParameterName, filePath, newLine, contentType)
            MsgBox(header)
            Dim headerBytes As Byte() = Encoding.UTF8.GetBytes(header.ToString)
            requestStream.Write(headerBytes, 0, header.Length)

            Using fileStream As New IO.FileStream(filePath, IO.FileMode.Open, IO.FileAccess.Read)
                Dim buffer(4096) As Byte
                Dim bytesRead As Int32 = fileStream.Read(buffer, 0, buffer.Length)
                Do While (bytesRead > 0)
                    requestStream.Write(buffer, 0, bytesRead)
                    bytesRead = fileStream.Read(buffer, 0, buffer.Length)
                Loop

            End Using
            Dim trailer As Byte() = Encoding.ASCII.GetBytes(newLine & "--" + boundary + "--" & newLine)
            requestStream.Write(trailer, 0, trailer.Length)
        End Using

        Dim response As Net.WebResponse = Nothing

        Try
            response = request.GetResponse()
            Using responseStream As IO.Stream = response.GetResponseStream()
                Using responseReader As New IO.StreamReader(responseStream)
                    Dim responseText = responseReader.ReadToEnd()
                    MsgBox(responseText)
                End Using
            End Using
        Catch exception As Net.WebException
            response = exception.Response
            If (response IsNot Nothing) Then
                Using reader As New IO.StreamReader(response.GetResponseStream())
                    Dim responseText = reader.ReadToEnd()
                    Diagnostics.Debug.Write(responseText)
                End Using
                response.Close()
            End If
        Finally
            process.Start(uri)
            request = Nothing
        End Try

    End Sub

End Class

PHP代码是：

<?php


$target_dir = "C:\images";
print_r($_FILES);
//print_r($_POST);
print_r($_REQUEST);

$target_file = $_FILES['file']['tmp_name'];

 if(move_uploaded_file($target_file, $target_dir. DIRECTORY_SEPARATOR . $_FILES['file']['name']))
 {
 echo "Success";
 }
 else
 {
 echo "Failure";
 }

?>

我很感激有任何帮助。

祝你好运！

尝试将.zip文件从VB .NET上传到服务器时出错

0 个答案: