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
€˛NickHK€¯ ezt Ć*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
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