TīmeklisLine 2: This line displays the Save As dialog box. ResultVar = Application.GetSaveAsFilename () Line 3: This line first checks if the user has entered a filename and there are no errors. If so, then it takes the entire path of the file, along with the filename the user specified, and displays it in a message box. http://duoduokou.com/excel/40873886512152360223.html
Default file selection using Application.GetOpenFilename
Tīmeklis2024. gada 25. jūl. · The macro: Public Sub ExportSheetsToCSV () Dim wsExport As Worksheet Dim wbkExport As Workbook For Each wsExport In Worksheets wsExport.Select nm = wsExport.Name If Not IsActiveSheetEmpty () Then ActiveSheet.SaveAs fileName:="H:\CSV_Split_Exports\" & nm, FileFormat:=xlCSV … Tīmeklis2024. gada 12. marts · Here's the current code: Code: Application.GetOpenFilename (FileFilter:="Microsoft Excel Workbooks (*.xls;*.xlsx;*.xlsm)*.xls;*.xlsx;*.xlsm", Title:="Open Database File", MultiSelect:=False) Gives me Method GetOpenFileName of object Application failed (error 1004). I'm sure it is not far from working. Excel Facts alava lascaray
excel - VBA - FileTo-Open a .csv file - Stack Overflow
Tīmeklis2024. gada 4. okt. · – Santhosh Sep 19, 2024 at 8:30 you still have not said which line causes the error. the format of the two files is not the same. single-step through your code and pay close attention to the value of StrData (0) after the strData () = Split (MyData, vbCrLf) line, and you may discover the reason – jsotola Sep 19, 2024 at … Tīmeklis2024. gada 7. janv. · The GetOpenFilename method displays the familiar Open dialog box (a dead ringer for the dialog box Excel displays when you choose File → Open → Browse). The GetOpenFilename method doesn’t actually open the specified file. This method simply returns the user-selected filename as a string. Then you can write … http://k1simplify.com/vba/tipsleaf/leaf12.html alava international