Opening *.txt by code with varible path
 

Hello to all,


I´m triying to open a *.txt file in a variable path and with a
variable name.


C:\.....\year\weekXX\

After opening the select dialog box I select the *.txt file and convert
with comma delimiters.



-year- is the value of a cell in worksheet
-weekXX- is the value of a cell in worksheet


Thanks for your help

Opening *.txt by code with varible path
 

Take a look at GetOpenFilename in VBA help.

--
HTH

Bob Phillips

"carloshernandezy" wrote in message
ups.com...
Hello to all,


I´m triying to open a *.txt file in a variable path and with a
variable name.


C:\.....\year\weekXX\

After opening the select dialog box I select the *.txt file and convert
with comma delimiters.



-year- is the value of a cell in worksheet
-weekXX- is the value of a cell in worksheet


Thanks for your help

Opening *.txt by code with varible path
 

Hello Bob,
I´ve been triying a code like this and it returns an error "13" in
line


Sub TXT_OPEN()

Dim varDir, varYear, varWeek As String
Dim ws As Workbook
Dim RutaArchivo As String
Set ws = ActiveWorkbook

Application.DisplayAlerts = False

varDir = "C:\....\VB\"
varYear = Range("L8")
varWeek = Range("L7")
RutaArchivo = varYear & " \ " & varWeek


ChDir " C:\....\VB\ varYear & " \ " & varWeek "

archivoAAbrir = Application.GetOpenFilename("Archivos de texto (*.txt),
*.txt")


If archivoAAbrir < False Then


ChDir " C:\....\VB\ "
Workbooks.OpenText Filename:=archivoAAbrir, Origin:= _
xlWindows, StartRow:=1, DataType:=xlDelimited, TextQualifier:=
_
xlDoubleQuote, ConsecutiveDelimiter:=False, Tab:=True,
Semicolon:=False, _
Comma:=True, Space:=False, Other:=True, OtherChar:="|",
FieldInfo:= _
Array(Array(1, 1), Array(2, 1), Array(3, 1), Array(4, 1),
Array(5, 1), Array(6, 1), Array(7 _
, 1), Array(8, 1))
End If


End Sub

Opening *.txt by code with varible path
 

You can't use "C:\.....\

you have to put in the path explicitly.

--
Regards,
Tom Ogilvy

"carloshernandezy" wrote in message
ups.com...
Hello Bob,
I´ve been triying a code like this and it returns an error "13" in
line


Sub TXT_OPEN()

Dim varDir, varYear, varWeek As String
Dim ws As Workbook
Dim RutaArchivo As String
Set ws = ActiveWorkbook

Application.DisplayAlerts = False

varDir = "C:\....\VB\"
varYear = Range("L8")
varWeek = Range("L7")
RutaArchivo = varYear & " \ " & varWeek


ChDir " C:\....\VB\ varYear & " \ " & varWeek "

archivoAAbrir = Application.GetOpenFilename("Archivos de texto (*.txt),
*.txt")


If archivoAAbrir < False Then


ChDir " C:\....\VB\ "
Workbooks.OpenText Filename:=archivoAAbrir, Origin:= _
xlWindows, StartRow:=1, DataType:=xlDelimited, TextQualifier:=
_
xlDoubleQuote, ConsecutiveDelimiter:=False, Tab:=True,
Semicolon:=False, _
Comma:=True, Space:=False, Other:=True, OtherChar:="|",
FieldInfo:= _
Array(Array(1, 1), Array(2, 1), Array(3, 1), Array(4, 1),
Array(5, 1), Array(6, 1), Array(7 _
, 1), Array(8, 1))
End If


End Sub

Opening *.txt by code with varible path
 

Tom I Know it´s only an example, it is complete in the original code.

Thanks for your advice.

Regards,

Opening *.txt by code with varible path
 

so this is your problem line?

ChDir " C:\....\VB\ varYear & " \ " & varWeek "

Unless you just omitted and added double quotes arbitrarily as examples,
then


ChDir "C:\....\VB\" & varYear & " \ " & varWeek

but since you did this

RutaArchivo = varYear & " \ " & varWeek


Why not just do

ChDir "C:\....\VB\" & RutaArchivo



--
Regards,
Tom Ogilvy



"carloshernandezy" wrote in message
oups.com...
Tom I Know it´s only an example, it is complete in the original code.

Thanks for your advice.

Regards,