XL2003 open .CSV problem
 

Hi All,

Earlier I used XL2000, and I had a macro opening a .CSV file:
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv"
worked correctly, it placed the semicolon-separated data into different
columns (I use a Hungarian version where list separator is semicolon).

Recently I had to upgrade to XL2003 (because I couldn't go without some
feature existing only in XL2003), and this statement doesn't work correctly
in XL2003: it opens the .csv file, but does NOT place the semicolon-separated
data into different columns.

I tried
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=4
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=6, delimiter:=";"
with the same result.

Can anybody help?

Regards,
Stefi

XL2003 open .CSV problem
 

Stefi,
Excel takes the delimiter from the Windows setting Regional
OptionsNumbersList Separator.
Set that to a semi colon or use the DataGet external dataImport Text file
menu.

NickHK

"Stefi" wrote in message
...
Hi All,

Earlier I used XL2000, and I had a macro opening a .CSV file:
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv"
worked correctly, it placed the semicolon-separated data into different
columns (I use a Hungarian version where list separator is semicolon).

Recently I had to upgrade to XL2003 (because I couldn't go without some
feature existing only in XL2003), and this statement doesn't work
correctly
in XL2003: it opens the .csv file, but does NOT place the
semicolon-separated
data into different columns.

I tried
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=4
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=6,
delimiter:=";"
with the same result.

Can anybody help?

Regards,
Stefi

XL2003 open .CSV problem
 

How about:

Workbooks.OpenText Filename:="C:\Myfolder\2005_12.csv",
DataType:=xlDelimited, Semicolon:=True

(I just got this by recording a macro, opening the file, using the import
wizard with a semicolon and then cutting out the needless arguments)

"Stefi" wrote:

Hi All,

Earlier I used XL2000, and I had a macro opening a .CSV file:
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv"
worked correctly, it placed the semicolon-separated data into different
columns (I use a Hungarian version where list separator is semicolon).

Recently I had to upgrade to XL2003 (because I couldn't go without some
feature existing only in XL2003), and this statement doesn't work correctly
in XL2003: it opens the .csv file, but does NOT place the semicolon-separated
data into different columns.

I tried
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=4
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=6, delimiter:=";"
with the same result.

Can anybody help?

Regards,
Stefi

XL2003 open .CSV problem
 

Hi Nick,
Thanks for your reply!
I checked my Windows setting Regional OptionsNumbersList Separator, it is
a semicolon.
I tried DataGet external dataImport Text file, it gave a worksheet-like
appearance (data in separate columns), but in a noname new workbook, not as
2005_12.csv workbook as in XL2000 which kept its .csv structure when saved,
and recorded as a macro it gave a very long list of statements instead of one
simple line.

The funny thing is that opening the .csv file manually gives the expected
correct sheet, and recording it as a macro generates this simple line:
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv"
but running this macro gives the incorrect result mentioned earlier.

Can you explain me why was it necessary to make worse what was good?

Regards,
Stefi


€žNickHK€� ezt Ã*rta:

Stefi,
Excel takes the delimiter from the Windows setting Regional
OptionsNumbersList Separator.
Set that to a semi colon or use the DataGet external dataImport Text file
menu.

NickHK

"Stefi" wrote in message
...
Hi All,

Earlier I used XL2000, and I had a macro opening a .CSV file:
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv"
worked correctly, it placed the semicolon-separated data into different
columns (I use a Hungarian version where list separator is semicolon).

Recently I had to upgrade to XL2003 (because I couldn't go without some
feature existing only in XL2003), and this statement doesn't work
correctly
in XL2003: it opens the .csv file, but does NOT place the
semicolon-separated
data into different columns.

I tried
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=4
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=6,
delimiter:=";"
with the same result.

Can anybody help?

Regards,
Stefi

XL2003 open .CSV problem
 

Thanks Martin,
Your idea put me on the right track, but the Opentext method worked only
when I changed the file extension from .csv to .txt!

My questions still obtain: why is not XL2003 compatible with XL2000?

Regards,
Stefi


€žMartin€� ezt Ã*rta:

How about:

Workbooks.OpenText Filename:="C:\Myfolder\2005_12.csv",
DataType:=xlDelimited, Semicolon:=True

(I just got this by recording a macro, opening the file, using the import
wizard with a semicolon and then cutting out the needless arguments)

"Stefi" wrote:

Hi All,

Earlier I used XL2000, and I had a macro opening a .CSV file:
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv"
worked correctly, it placed the semicolon-separated data into different
columns (I use a Hungarian version where list separator is semicolon).

Recently I had to upgrade to XL2003 (because I couldn't go without some
feature existing only in XL2003), and this statement doesn't work correctly
in XL2003: it opens the .csv file, but does NOT place the semicolon-separated
data into different columns.

I tried
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=4
Workbooks.Open Filename:="C:\Myfolder\2005_12.csv", Format:=6, delimiter:=";"
with the same result.

Can anybody help?

Regards,
Stefi