S
Stefi
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
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