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variable 'open' command
Hi Guys
I have the following problem: I have to copy data from one excelfile (1.xls) to another (summery.xls). No problem so far. it looks like Workbooks.Open ("c:\[...]\1.xls") Windows("1.xls").Activate But now the path and the file itself changes! i need some kind of query to tell to program which file it should use. I couldn't find anything in the helpfiles which are not helpful at all, btw. Help is very much appreciated! Walt |
#2
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variable 'open' command
Does this mean that C:\...\1.xls can change name and location?
If yes, you can ask the user to just click on the file using the File|open dialog. Option Explicit Sub testme() Dim myFileName As Variant Dim myWkbk As Workbook myFileName = Application.GetOpenFilename("Excel files, *.xls") If myFileName = False Then 'user hit cancel Exit Sub '?? End If Set myWkbk = Workbooks.Open(Filename:=myFileName) 'Now you can refer to the variable mywkbk instead of the actual name. mywkbk.worksheets(1).range("A1").value = date mywkbk.close savechanges:=true End Sub Walt wrote: Hi Guys I have the following problem: I have to copy data from one excelfile (1.xls) to another (summery.xls). No problem so far. it looks like Workbooks.Open ("c:\[...]\1.xls") Windows("1.xls").Activate But now the path and the file itself changes! i need some kind of query to tell to program which file it should use. I couldn't find anything in the helpfiles which are not helpful at all, btw. Help is very much appreciated! Walt -- Dave Peterson |
#3
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variable 'open' command
Thx, Dave. That helped a lot! But there is still a problem. If I try
(as Henry and I thought would be possible) Workbooks("myWkbk").Activate i get an error message: "Run-time error '9': Subscript out of range" Any clue for that one? TIA Walt Dave Peterson wrote in message ... Does this mean that C:\...\1.xls can change name and location? If yes, you can ask the user to just click on the file using the File|open dialog. Option Explicit Sub testme() Dim myFileName As Variant Dim myWkbk As Workbook myFileName = Application.GetOpenFilename("Excel files, *.xls") If myFileName = False Then 'user hit cancel Exit Sub '?? End If Set myWkbk = Workbooks.Open(Filename:=myFileName) 'Now you can refer to the variable mywkbk instead of the actual name. mywkbk.worksheets(1).range("A1").value = date mywkbk.close savechanges:=true End Sub Walt wrote: Hi Guys I have the following problem: I have to copy data from one excelfile (1.xls) to another (summery.xls). No problem so far. it looks like Workbooks.Open ("c:\[...]\1.xls") Windows("1.xls").Activate But now the path and the file itself changes! i need some kind of query to tell to program which file it should use. I couldn't find anything in the helpfiles which are not helpful at all, btw. Help is very much appreciated! Walt |
#4
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variable 'open' command
Hi Walt,
Your problem statement is a bit vague. How is the program supposed to determine the file name? David Gray P6 Consulting http://www.p6c.com You are more important than any technology we may employ. -----Original Message----- But now the path and the file itself changes! i need some kind of query to tell to program which file it should use. I couldn't find anything in the helpfiles which are not helpful at all, btw. |
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