Opening semicolon delimited CSV file

Question

How does one open a semicolon delimited CSV file with VBA in Excel 2000?

Sample data

An ID;TEST20090222
A Name;Firstname Surname
A D         


        

Answer 1

I find that this works for me in Excel 2000:

Workbooks.OpenText filename:=myFilename, _
    DataType:=xlDelimited, Semicolon:=True

Answer 2

Here's the OpenText method from Excel 2000:

OpenText Method

Loads and parses a text file as a new workbook with a single sheet that contains the parsed text-file data.

Syntax

expression.OpenText(Filename, Origin, StartRow, DataType, TextQualifier, ConsecutiveDelimiter, Tab, Semicolon, Comma, Space, Other, OtherChar, FieldInfo, DecimalSeparator, ThousandsSeparator)

source

and here's the Excel 2003 version:

OpenText Method [Excel 2003 VBA Language Reference]

Loads and parses a text file as a new workbook with a single sheet that contains the parsed text-file data.

expression.OpenText(FileName, Origin, StartRow, DataType, TextQualifier, ConsecutiveDelimiter, Tab, Semicolon, Comma, Space, Other, OtherChar, FieldInfo, TextVisualLayout, DecimalSeparator, ThousandsSeparator, TrailingMinusNumbers, Local)

source

so Local was indeed a new parameter for Excel 2003 and won't work in Excel 2000

No idea as to the cause of the erroneous behaviour. The Local parameter is defined as:

Local Optional Variant. Specify True if regional settings of the machine should be used for separators, numbers and data formatting.

You might want to double-check the regional settings on the Excel 2000 PC and check to see if there is anything which may cause the data to be wrongly interpreted. Also, try explicitly specifying the DecimalSeparator and ThousandsSeparator parameters on the Excel 2000 method and see if that helps