【发布时间】:2019-12-28 22:04:21
【问题描述】:
我有一些 vba 来创建一个工作表名称为“SheetName”的新工作簿,然后将数据传输到新工作表并以文件名另存为并保存为 CSV 格式 文件名复制了我之前声明的表单名称,这是我不想要的。
Private Sub TransferData()
'---Submit Button Code - Transfers data to new workbook, renames and saves to folder.
Dim SheetName As String
Dim SavePath As String
Dim FileName As String
SheetName = "IATData"
SavePath = "savedfolderpath"
FileName = IAT.SaveDateBox.Value & IAT.SaveTimeBox.Value & IAT.SaveAssessorBRIDBox & IAT.BRIDBox
'Create new workbook for saved CSV
With Application
.SheetsInNewWorkbook = 1
.Workbooks.Add
.Sheets(1).Name = SheetName
End With
'Turn off Auto-Calculate
Application.Calculation = xlCalculationManual
'Turn off Alert Displays
Application.DisplayAlerts = False
'Determine Empty Row
emptyRow = WorksheetFunction.CountA(Range("A:A")) + 1
'Transfer the data to the master
'---Front Page---
Cells(emptyRow, 1).Value = IAT.BRIDBox.Value
Cells(emptyRow, 2).Value = IAT.AgentNameBox.Value
'Save new workbook etc.
ActiveWorkbook.SaveAs FileName:=SavePath & FileName, FileFormat:=xlCSV, CreateBackup:=False
ActiveWorkbook.Close
Application.DisplayAlerts = True
Workbooks("IAT Form.xlsm").Activate
MsgBox "Details sent to data folder under: " & FileName
End Sub
希望新工作簿具有工作表名称“IATData”和由上述单元格引用组成的文件名,但工作表仍被命名为文件名。
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