我想使用VBA以编程方式刷新Word模板报表中的Excel表格。这些表在Matlab的模板Excel文件中写入多张表。文件结构如下所示:
代码必须检查文件夹结构,看它是否将Excel文件从最新的文件夹中拉出来。如果是,它只会刷新所有细胞。如果没有,则必须删除所有表格,并从之前的表格中拉出新表格。我不确定下面的星号之间的代码。任何帮助将不胜感激。
Sub LinkToCurrentTableFolder()
'Get current folder by date
Dim clientTablesPath As Variant
filePath = ActiveDocument.Path & "\ClientTables\"
Set fso = CreateObject("Scripting.FileSystemObject")
Set fld = fso.GetFolder(filePath)
Dim currentFolder As Variant: currentFolder = ""
For Each sf In fld.SUBFOLDERS
'Look at name and get current date
If currentFolder = "" Then
currentFolder = sf.Path
ElseIf sf.Path > currentFolder Then
currentFolder = sf.Path
End If
Next
'***
'Debug: display current Excel folder path
'MsgBox (currentFolder)
If currentPath = currentFolder Then
'Loop through all tables in document and refresh
'If path is not current delete current table
Dim tbTemp As Table
Dim cellTemp As Cell
For Each tbTemp In ActiveDocument.Tables
For Each cellTemp In tbTemp.Range.Cells
cellTemp.Range.Fields.Update
Next
Next
Else
'Locate same file name in new folder
shpName = .LinkFormat.SourceName
NewPath = currentFolder & "\" & shpName
'Delete existing table (???) Not sure
.Delete
'Create new table (???) Not sure - must be from same location and same size as previous one
Selection.Table.AddOLEObject ClassType:=cType, FileName:=NewPath, LinkToFile:=True, DisplayAsIcon:=False
End If
'***
End Sub
编辑 - 复制和粘贴完成如下所示:
答案 0 :(得分:2)
我找到了答案here - 此代码会询问新Excel文件的位置,并更新Excel链接表格的所有字段代码。
该链接的代码如下所示。
Public Sub changeSource()
Dim dlgSelectFile As FileDialog 'FileDialog object '
Dim thisField As Field
Dim selectedFile As Variant
'must be Variant to contain filepath of selected item
Dim newFile As Variant
Dim fieldCount As Integer '
Dim x As Long
On Error GoTo LinkError
'create FileDialog object as File Picker dialog box
Set dlgSelectFile = Application.FileDialog(FileDialogType:=msoFileDialogFilePicker)
With dlgSelectFile
.Filters.Clear 'clear filters
.Filters.Add "Microsoft Excel Files", "*.xls, *.xlsb, *.xlsm, *.xlsx" 'filter for o nly Excel files
'use Show method to display File Picker dialog box and return user's action
If .Show = -1 Then
'step through each string in the FileDialogSelectedItems collection
For Each selectedFile In .SelectedItems
newFile = selectedFile 'gets new filepath
Next selectedFile
Else 'user clicked cancel
Exit Sub
End If
End With
Set dlgSelectFile = Nothing
'update fields
With ActiveDocument
fieldCount = .Fields.Count
For x = 1 To fieldCount
With .Fields(x)
'Debug.Print x '
Debug.Print .Type
If .Type = 56 Then
'only update Excel links. Type 56 is an excel link
.LinkFormat.SourceFullName = newFile '
.Update
.LinkFormat.AutoUpdate = False
DoEvents
End If
End With
Next x
End With
MsgBox "Source data has been successfully imported."
Exit Sub
LinkError:
Select Case Err.Number
Case 5391 'could not find associated Range Name
MsgBox "Could not find the associated Excel Range Name " & _
"for one or more links in this document. " & _
"Please be sure that you have selected a valid " & _
"Quote Submission input file.", vbCritical
Case Else
MsgBox "Error " & Err.Number & ": " & Err.Description, vbCritical
End Select
End Sub