Visual Basic-将文件上载到PHP网页

我正在尝试将一个图像文件上载到web服务器上的PHP文件

在VB.NET上->

My.Computer.Network.UploadFile(tempImageLocation, "website.com/upload.php")

tempImageLocation是硬盘上映像所在的位置。图像位于我指定的硬盘驱动器上

关于PHP->

$image = $_FILES['uploads']['name'];

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我不明白，因为它正在加载页面-但PHP在“上载”下找不到文件。

这里有一个快速而肮脏的教程供您使用：

“uploads”只是表单元素的名称属性值：

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或者换句话说，这是通过$_filesglobal访问的POST变量名。

如果不设置字段名，则可以使用此字段保存上载的文件

$file = array_shift($_FILES);
move_uploaded_file($file['tmp_name'], '/path/to/new/location/'.$file['name']);

请看一些答案。PHP要求使用POST方法上载的文件，这些文件通常在从web表单上载时由浏览器设置，但可以在VB中使用设置

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至于PHP端，在使用

$image=$\u文件['uploads']['name']上传后，您将无法立即找到该文件。PHP使用一个临时文件名存储上传，该文件名可通过
$\u FILES['uploads']['tmp\u name']
变量访问，使用
move\u uploadd\u file（）
是将上传从临时存储转移到永久上传目录的标准方法。在我搜索同一个问题时，谷歌把我带到了这里。感谢大家，它给了我这个想法，我对PHP有了一点了解，就实现了。我知道这是一个老问题，但我还是要和大家分享我的代码，这样将来它可以帮助人们

VB:

PHP:

不要忘了给上传文件夹适当的权限。
下面是使用Visual Basic上传文件的完整示例，在服务器端PHP（Rest API）
我知道，它已经过时了。。但我有一个解决方案：

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

    Dim myFile As New FileInfo(filePath)
    Dim sizeInBytes As Long = myFile.Length

    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)
        header = header & vbNewLine & "Content-Length: " & sizeInBytes.ToString & vbNewLine
        header = header & "Expect: 100-continue" & vbNewLine & vbNewLine

        'MsgBox(header)
        Debug.Print(header)

        Dim headerBytes As Byte() = Encoding.UTF8.GetBytes(header)
        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)
        requestStream.Close()
    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()
                Debug.Print(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
        request = Nothing
    End Try
End Sub

使用：

HttpUploadFile("https://www.yousite.com/ws/upload.php?option1=sss&options2=12121", FULL_FILE_NAME_PATH_IN_YOUR_PC, "files", "multipart/form-data")

我在一个我不记得的网站上复制了一些代码。
我只使用了这两行代码：

header=header&vbNewLine&“内容长度：”&sizeInBytes.ToString&vbNewLine
header=header&vbNewLine&“Expect:100 continue”&vbNewLine

希望得到帮助。
这是我的示例服务器php文件：

<?php
// write to a log file so you know it's working
$msg = $_POST['w'];
$logfile= 'data.txt';

$fp = fopen($logfile, "a");
fwrite($fp, $msg);
fclose($fp);

$file = array_shift($_FILES);
move_uploaded_file($file['tmp_name'], '/MAMP/htdocs/test/'.$file['name']);

?>

哈哈。有趣的是，之所以需要通过Visual Basic上传，是因为我的应用程序正在制作图像，需要上传，而且需要无缝上传。我不希望我的用户必须找到图像本身才能上传它！：太感谢你了！虽然我不再需要它了，但我非常感谢你花时间帮助我回答这个问题。
HttpUploadFile("https://www.yousite.com/ws/upload.php?option1=sss&options2=12121", FULL_FILE_NAME_PATH_IN_YOUR_PC, "files", "multipart/form-data")

<?php
// write to a log file so you know it's working
$msg = $_POST['w'];
$logfile= 'data.txt';

$fp = fopen($logfile, "a");
fwrite($fp, $msg);
fclose($fp);

$file = array_shift($_FILES);
move_uploaded_file($file['tmp_name'], '/MAMP/htdocs/test/'.$file['name']);

?>

HttpUploadFile("http://localhost/test/test.php?w=hello&options2=12121", "C:\temp\bahamas.mp3", "files", "multipart/form-data")