Write Procedure With Logic Test To Open File From Given Directory

[Carl Bowman]

I need to write a procedure (not sure if this is what you call it) to cause
the file "data.xls" to be opened from path "C:\data\FEB" if the month is
February (evidenced by the text "FEB" in cell L45 in worksheet "Process") and
a file with the same name to be opened from path "C:\data\MAR" if the month
is March (evidenced by the text "MAR" in cell L45 of worksheet "Process").
This may be simple for some but I don't even know where to start. Thanks!

[Markus Scheible]

Hi Carl,

>the file "data.xls" to be opened from path "C:\data\FEB"
if the month is
>February (evidenced by the text "FEB" in cell L45 in
worksheet "Process") and


Try:

workbooks.Open FileName:="C:\data\" & Sheets
("Process").Range("L45").Value & "\data.xls"

Best

Markus

[David Fam?]

OK,
if I've undestood, you want to open files with the same names and
different paths variable on a cell value. If this is correct, do as
follows:

Create a command button (from the Control Toolbox) and (in the Design
Mode - always selectable in the Control Tollbox Taskbar) double click
it. A macro screen will popup with a couple of lines written:

>> Private Sub CommandButton1_Click()
>> ' Here write the following code
>> End Sub

The following code will create from cell L45 the right path and will
open the file (without ">>"):

>> ' Starts Here
>>
>> dim sPath as string
>> dim sPartialPath as string
>> dim sFileName as string
>>
>> sFileName = "data.xls"
>> sPartialPath = worksheets("Process").cells(45,12).value
>> sPath = "C:\data\" & sPartialPath & "\" & sFileName
>>
>> Workbooks.open(sPath)
>> ' Finishes Here

Now, saving and then pushing the button (NOT in Design Mode) the
desired file should be opened.

Hope it Helps,
David

[Carl Bowman]

Thanks a million Markus! This worked great and has opened up a whole new
world to me!

"Markus Scheible" wrote:

> Hi Carl,
>
> >the file "data.xls" to be opened from path "C:\data\FEB"
> if the month is
> >February (evidenced by the text "FEB" in cell L45 in
> worksheet "Process") and
>
>
> Try:
>
> workbooks.Open FileName:="C:\data\" & Sheets
> ("Process").Range("L45").Value & "\data.xls"
>
> Best
>
> Markus
>

[Carl Bowman]

Thanks, David. I will need to study this more. I was able to use the other
suggestion and it worked just fine. I still see some use for what you are
saying.

"David Fam?" wrote:

> OK,
> if I've undestood, you want to open files with the same names and
> different paths variable on a cell value. If this is correct, do as
> follows:
>
> Create a command button (from the Control Toolbox) and (in the Design
> Mode - always selectable in the Control Tollbox Taskbar) double click
> it. A macro screen will popup with a couple of lines written:
>
> >> Private Sub CommandButton1_Click()
> >> ' Here write the following code
> >> End Sub
>
> The following code will create from cell L45 the right path and will
> open the file (without ">>"):
>
> >> ' Starts Here
> >>
> >> dim sPath as string
> >> dim sPartialPath as string
> >> dim sFileName as string
> >>
> >> sFileName = "data.xls"
> >> sPartialPath = worksheets("Process").cells(45,12).value
> >> sPath = "C:\data\" & sPartialPath & "\" & sFileName
> >>
> >> Workbooks.open(sPath)
> >> ' Finishes Here
>
> Now, saving and then pushing the button (NOT in Design Mode) the
> desired file should be opened.
>
> Hope it Helps,
> David
>