这个线程现在可能已经死了,但是如果您正在使用数据库中的表单并且用户需要将文件附加到以具有唯一标识号的表单中显示的特定记录,那么这绝对是可能的,但您必须这样做在用 .NET 编写的外部应用程序中,我可以为您提供必要的代码来帮助您入门,vb.net 与 VBA 非常相似。
您需要做的是创建一个 Windows 窗体项目并添加对 Microsoft 访问核心 dll 的引用,并从 nugget 下载 google drive api 的 nugget 包。
Imports Google
Imports Google.Apis.Services
Imports Google.Apis.Drive.v2
Imports Google.Apis.Auth.OAuth2
Imports Google.Apis.Drive.v2.Data
Imports System.Threading
Public Class GoogleDriveAuth
Public Shared Function GetAuthentication() As DriveService
Dim ClientIDString As String = "Your Client ID"
Dim ClientSecretString As String = "Your Client Secret"
Dim ApplicationNameString As String = "Your Application Name"
Dim secrets = New ClientSecrets()
secrets.ClientId = ClientIDString
secrets.ClientSecret = ClientSecretString
Dim scope = New List(Of String)
scope.Add(DriveService.Scope.Drive)
Dim credential = GoogleWebAuthorizationBroker.AuthorizeAsync(secrets, scope, "user", CancellationToken.None).Result()
Dim initializer = New BaseClientService.Initializer
initializer.HttpClientInitializer = credential
initializer.ApplicationName = ApplicationNameString
Dim Service = New DriveService(initializer)
Return Service
End Function
End Class
此代码将授权您的驱动器服务,然后您在导入下创建一个公共共享服务作为 DriveService 可以从任何子或函数中使用,然后在您的表单加载事件中调用此函数,例如
服务 = GoogleDriveAuth.GetAuthentication
将对您的项目的引用添加到 Microsoft Access 12.0 对象库或您拥有的任何版本
然后这段代码将查看您想要从中获取记录值的表单并将文件上传到您选择的文件夹
Private Sub UploadAttachments()
Dim NumberExtracted As String
Dim oAccess As Microsoft.Office.Interop.Access.Application = Nothing
Dim connectedToAccess As Boolean = False
Dim SelectedFolderIdent As String = "Your Upload Folder ID"
Dim CreatedFolderIdent As String
Dim tryToConnect As Boolean = True
Dim oForm As Microsoft.Office.Interop.Access.Form
Dim oCtls As Microsoft.Office.Interop.Access.Controls
Dim oCtl As Microsoft.Office.Interop.Access.Control
Dim sForm As String 'name of form to show
sForm = "Your Form Name"
Try
While tryToConnect
Try
' See if can connect to a running Access instance
oAccess = CType(Marshal.GetActiveObject("Access.Application"), Microsoft.Office.Interop.Access.Application)
connectedToAccess = True
Catch ex As Exception
Try
' If couldn't connect to running instance of Access try to start a running Access instance And get an updated version of the database
oAccess = CType(CreateObject("Access.Application"), Microsoft.Office.Interop.Access.Application)
oAccess.Visible = True
oAccess.OpenCurrentDatabase("Your Database Path", False)
connectedToAccess = True
Catch ex2 As Exception
Dim res As DialogResult = MessageBox.Show("COULD NOT CONNECT TO OR START THE DATABASE" & vbNewLine & ex2.Message, "Warning", MessageBoxButtons.AbortRetryIgnore, MessageBoxIcon.Warning)
If res = System.Windows.Forms.DialogResult.Abort Then
Exit Sub
End If
If res = System.Windows.Forms.DialogResult.Ignore Then
tryToConnect = False
End If
End Try
End Try
' We have connected successfully; stop trying
tryToConnect = False
End While
' Start a new instance of Access for Automation:
' Make sure Access is visible:
If Not oAccess.Visible Then oAccess.Visible = True
' For Each oForm In oAccess.Forms
' oAccess.DoCmd.Close(ObjectType:=Microsoft.Office.Interop.Access.AcObjectType.acForm, ObjectName:=oForm.Name, Save:=Microsoft.Office.Interop.Access.AcCloseSave.acSaveNo)
' Next
' If Not oForm Is Nothing Then
' System.Runtime.InteropServices.Marshal.ReleaseComObject(oForm)
' End If
' oForm = Nothing
' Select the form name in the database window and give focus
' to the database window:
' oAccess.DoCmd.SelectObject(ObjectType:=Microsoft.Office.Interop.Access.AcObjectType.acForm, ObjectName:=sForm, InDatabaseWindow:=True)
' Show the form:
' oAccess.DoCmd.OpenForm(FormName:=sForm, View:=Microsoft.Office.Interop.Access.AcFormView.acNormal)
' Use Controls collection to edit the form:
oForm = oAccess.Forms(sForm)
oCtls = oForm.Controls
oCtl = oCtls.Item("The Name Of The Control Where The Id Number Is On The Form")
oCtl.Enabled = True
' oCtl.SetFocus()
NumberExtracted = oCtl.Value
System.Runtime.InteropServices.Marshal.ReleaseComObject(oCtl)
oCtl = Nothing
' Hide the Database Window:
' oAccess.DoCmd.SelectObject(ObjectType:=Microsoft.Office.Interop.Access.AcObjectType.acForm, ObjectName:=sForm, InDatabaseWindow:=True)
' oAccess.RunCommand(Command:=Microsoft.Office.Interop.Access.AcCommand.acCmdWindowHide)
' Set focus back to the form:
' oForm.SetFocus()
' Release Controls and Form objects:
System.Runtime.InteropServices.Marshal.ReleaseComObject(oCtls)
oCtls = Nothing
System.Runtime.InteropServices.Marshal.ReleaseComObject(oForm)
oForm = Nothing
' Release Application object and allow Access to be closed by user:
If Not oAccess.UserControl Then oAccess.UserControl = True
System.Runtime.InteropServices.Marshal.ReleaseComObject(oAccess)
oAccess = Nothing
If NumberExtracted = Nothing Then
MsgBox("The Number Could Not Be Obtained From The Form" & vbNewLine & vbNewLine & "Please Ensure You Have The Form Open Before Trying To Upload")
Exit Sub
End If
If CheckForDuplicateFolder(SelectedFolderIdent, NumberExtracted + " - ATC") = True Then
CreatedFolderIdent = GetCreatedFolderID(NumberExtracted + " - ATC", SelectedFolderIdent)
DriveFilePickerUploader(CreatedFolderIdent)
Else
CreateNewDriveFolder(NumberExtracted + " - ATC", SelectedFolderIdent)
CreatedFolderIdent = GetCreatedFolderID(NumberExtracted + " - ATC", SelectedFolderIdent)
DriveFilePickerUploader(CreatedFolderIdent)
End If
Catch EX As Exception
MsgBox("The Number Could Not Be Obtained From The Form" & vbNewLine & vbNewLine & "Please Ensure You Have The Form Open Before Trying To Upload" & vbNewLine & vbNewLine & EX.Message)
Exit Sub
Finally
If Not oCtls Is Nothing Then
System.Runtime.InteropServices.Marshal.ReleaseComObject(oCtls)
oCtls = Nothing
End If
If Not oForm Is Nothing Then
System.Runtime.InteropServices.Marshal.ReleaseComObject(oForm)
oForm = Nothing
End If
If Not oAccess Is Nothing Then
System.Runtime.InteropServices.Marshal.ReleaseComObject(oAccess)
oAccess = Nothing
End If
End Try
End
End Sub
检查目标上传文件夹中的重复文件夹
Public Function CheckForDuplicateFolder(ByVal FolderID As String, ByVal NewFolderNameToCheck As String) As Boolean
Dim ResultToReturn As Boolean = False
Try
Dim request = Service.Files.List()
Dim requeststring As String = ("'" & FolderID & "' in parents And mimeType='application/vnd.google-apps.folder' And trashed=false")
request.Q = requeststring
Dim FileList = request.Execute()
For Each File In FileList.Items
If File.Title = NewFolderNameToCheck Then
ResultToReturn = True
End If
Next
Catch EX As Exception
MsgBox("THERE HAS BEEN AN ERROR" & EX.Message)
End Try
Return ResultToReturn
End Function
创建新的驱动器文件夹
Public Sub CreateNewDriveFolder(ByVal DirectoryName As String, ByVal ParentFolder As String)
Try
Dim body1 = New Google.Apis.Drive.v2.Data.File
body1.Title = DirectoryName
body1.Description = "Created By Automation"
body1.MimeType = "application/vnd.google-apps.folder"
body1.Parents = New List(Of ParentReference)() From {New ParentReference() With {.Id = ParentFolder}}
Dim file1 As Google.Apis.Drive.v2.Data.File = Service.Files.Insert(body1).Execute()
Catch EX As Exception
MsgBox("THERE HAS BEEN AN ERROR" & EX.Message)
End Try
End Sub
获取创建的文件夹 ID
Public Function GetCreatedFolderID(ByVal FolderName As String, ByVal FolderID As String) As String
Dim ParentFolder As String
Try
Dim request = Service.Files.List()
Dim requeststring As String = ("'" & FolderID & "' in parents And mimeType='application/vnd.google-apps.folder' And title='" & FolderName & "' And trashed=false")
request.Q = requeststring
Dim Parent = request.Execute()
ParentFolder = (Parent.Items(0).Id)
Catch EX As Exception
MsgBox("THERE HAS BEEN AN ERROR" & EX.Message)
End Try
Return ParentFolder
End Function
驱动文件选择器上传器将从文件对话框中选择的文件上传到新创建的文件夹
Public Sub DriveFilePickerUploader(ByVal ParentFolderID As String)
Try
ProgressBar1.Value = 0
Dim MimeTypeToUse As String
Dim dr As DialogResult = Me.OpenFileDialog1.ShowDialog()
If (dr = System.Windows.Forms.DialogResult.OK) Then
Dim file As String
Else : Exit Sub
End If
Dim i As Integer = 0
For Each file In OpenFileDialog1.FileNames
MimeTypeToUse = GetMimeType(file)
Dim filetitle As String = (OpenFileDialog1.SafeFileNames(i))
Dim body2 = New Google.Apis.Drive.v2.Data.File
body2.Title = filetitle
body2.Description = "J-T Auto File Uploader"
body2.MimeType = MimeTypeToUse
body2.Parents = New List(Of ParentReference)() From {New ParentReference() With {.Id = ParentFolderID}}
Dim byteArray = System.IO.File.ReadAllBytes(file)
Dim stream = New System.IO.MemoryStream(byteArray)
Dim request2 = Service.Files.Insert(body2, stream, MimeTypeToUse)
request2.Upload()
Next
Catch EX As Exception
MsgBox("THERE HAS BEEN AN ERROR" & EX.Message)
End Try
End Sub
获取上传文件的 Mime 类型
Public Shared Function GetMimeType(ByVal file As String) As String
Dim mime As String = Nothing
Dim MaxContent As Integer = CInt(New FileInfo(file).Length)
If MaxContent > 4096 Then
MaxContent = 4096
End If
Dim fs As New FileStream(file, FileMode.Open)
Dim buf(MaxContent) As Byte
fs.Read(buf, 0, MaxContent)
fs.Close()
Dim result As Integer = FindMimeFromData(IntPtr.Zero, file, buf, MaxContent, Nothing, 0, mime, 0)
Return mime
End Function
<DllImport("urlmon.dll", CharSet:=CharSet.Auto)> _
Private Shared Function FindMimeFromData( _
ByVal pBC As IntPtr, _
<MarshalAs(UnmanagedType.LPWStr)> _
ByVal pwzUrl As String, _
<MarshalAs(UnmanagedType.LPArray, ArraySubType:=UnmanagedType.I1, SizeParamIndex:=3)> ByVal _
pBuffer As Byte(), _
ByVal cbSize As Integer, _
<MarshalAs(UnmanagedType.LPWStr)> _
ByVal pwzMimeProposed As String, _
ByVal dwMimeFlags As Integer, _
<MarshalAs(UnmanagedType.LPWStr)> _
ByRef ppwzMimeOut As String, _
ByVal dwReserved As Integer) As Integer
End Function
希望这可以帮助您开始,我 100% 相信这是可以实现的,因为我已经为我的经理做了这件事。