如何在 VBA 中获取 excel 文件名/路径

Question

Say, I'm writing a VBA inside my excel file sample.xls . Now I want to get the full path of sample.xls in my VBA. How do I do it?

Answer 1

If you mean VBA, then you can use FullName, for example:

strFileFullName = ThisWorkbook.FullName

(updated as considered by the comments: the former used ActiveWorkbook.FullName could more likely be wrong, if other office files may be open(ed) and active. But in case you stored the macro in another file, as mentioned by user @user7296559 here , and really want the file name of the macro-using file, ActiveWorkbook could be the correct choice, if it is guaranteed to be active at execution time.)

Answer 2

this is a simple alternative that gives all responses, Fullname, Path, filename.

Dim FilePath, FileOnly, PathOnly As String

FilePath = ThisWorkbook.FullName
FileOnly = ThisWorkbook.Name
PathOnly = Left(FilePath, Len(FilePath) - Len(FileOnly))

Answer 3

   strScriptFullname = WScript.ScriptFullName 
   strScriptPath = Left(strScriptFullname, InStrRev(strScriptFullname,"\")) 

Answer 4

如果您只需要路径，这是最直接的方法：

PathOnly = ThisWorkbook.Path

Answer 5

ActiveWorkbook.FullName would be better I think, in case you have the VBA Macro stored in another Excel Workbook, but you want to get the details of the Excel you are editing, not where the Macro resides.

If they reside in the same file, then it does not matter, but if they are in different files, and you want the file where the Data is rather than where the Macro is, then ActiveWorkbook is the one to go for, because it deals with both scenarios.

Answer 6

if you need path only without file name:

ActiveWorkbook.Path

it would return D:\\Folder

if you need file path with file name also:

ActiveWorkbook.FullName

it would return D:\\Folder\\sample.xls

if you need file name only:

ActiveWorkbook.Name

it would return sample.xls

so if you want combine file path and file name to get full directory don't forget to add "" between. otherwise its simpler using .Path

Answer 7

There is a universal way to get this:

Function FileName() As String
    FileName = Mid(Application.Caption, 1, InStrRev(Application.Caption, "-") - 2)
End Function