Open CSV file via VBA (performance) - excel

Question

Obviously, this question has been asked many times. The normal procedure:
Workbooks.Open (ActiveWorkbook.Path & "\Test.csv")
won't parse the CSV correctly (having many rows in one cell)
Thanks to Lernkurve, I can use his function to get it right: Opening semicolon delimited CSV file
Sub ImportCSVFile(filepath As String)
Dim line As String
Dim arrayOfElements
Dim linenumber As Integer
Dim elementnumber As Integer
Dim element As Variant
linenumber = 0
elementnumber = 0
Open filepath For Input As #1 ' Open file for input
Do While Not EOF(1) ' Loop until end of file
linenumber = linenumber + 1
Line Input #1, line
arrayOfElements = Split(line, ";")
elementnumber = 0
For Each element In arrayOfElements
elementnumber = elementnumber + 1
Cells(linenumber, elementnumber).Value = element
Next
Loop
Close #1 ' Close file.
End Sub
This however is not fast (I have files with thousands of columns) and my question is:
Is there any native way to open CSV files in Excel with right parsing?

Answer 1

Workbooks.Open does work too.
Workbooks.Open ActiveWorkbook.Path & "\Temp.csv", Local:=True
this works/is needed because i use Excel in germany and excel does use "," to separate .csv by default because i use an english installation of windows. even if you use the code below excel forces the "," separator.
Workbooks.Open ActiveWorkbook.Path & "\Test.csv", , , 6, , , , , ";"
and Workbooks.Open ActiveWorkbook.Path & "\Temp.csv", , , 4 +variants of this do not work(!)
why do they even have the delimiter parameter if it is blocked by the Local parameter ?! this makes no sense at all. but now it works.

Answer 2

This function reads a CSV file of 15MB and copies its content into a sheet in about 3 secs.
What is probably taking a lot of time in your code is the fact that you copy data cell by cell instead of putting the whole content at once.
Option Explicit
Public Sub test()
copyDataFromCsvFileToSheet "C:\temp\test.csv", ",", "Sheet1"
End Sub
Private Sub copyDataFromCsvFileToSheet(parFileName As String, parDelimiter As String, parSheetName As String)
Dim data As Variant
data = getDataFromFile(parFileName, parDelimiter)
If Not isArrayEmpty(data) Then
With Sheets(parSheetName)
.Cells.ClearContents
.Cells(1, 1).Resize(UBound(data, 1), UBound(data, 2)) = data
End With
End If
End Sub
Public Function isArrayEmpty(parArray As Variant) As Boolean
'Returns false if not an array or dynamic array that has not been initialised (ReDim) or has been erased (Erase)
If IsArray(parArray) = False Then isArrayEmpty = True
On Error Resume Next
If UBound(parArray) < LBound(parArray) Then isArrayEmpty = True: Exit Function Else: isArrayEmpty = False
End Function
Private Function getDataFromFile(parFileName As String, parDelimiter As String, Optional parExcludeCharacter As String = "") As Variant
'parFileName is supposed to be a delimited file (csv...)
'parDelimiter is the delimiter, "," for example in a comma delimited file
'Returns an empty array if file is empty or can't be opened
'number of columns based on the line with the largest number of columns, not on the first line
'parExcludeCharacter: sometimes csv files have quotes around strings: "XXX" - if parExcludeCharacter = """" then removes the quotes
Dim locLinesList() As Variant
Dim locData As Variant
Dim i As Long
Dim j As Long
Dim locNumRows As Long
Dim locNumCols As Long
Dim fso As Variant
Dim ts As Variant
Const REDIM_STEP = 10000
Set fso = CreateObject("Scripting.FileSystemObject")
On Error GoTo error_open_file
Set ts = fso.OpenTextFile(parFileName)
On Error GoTo unhandled_error
'Counts the number of lines and the largest number of columns
ReDim locLinesList(1 To 1) As Variant
i = 0
Do While Not ts.AtEndOfStream
If i Mod REDIM_STEP = 0 Then
ReDim Preserve locLinesList(1 To UBound(locLinesList, 1) + REDIM_STEP) As Variant
End If
locLinesList(i + 1) = Split(ts.ReadLine, parDelimiter)
j = UBound(locLinesList(i + 1), 1) 'number of columns
If locNumCols < j Then locNumCols = j
i = i + 1
Loop
ts.Close
locNumRows = i
If locNumRows = 0 Then Exit Function 'Empty file
ReDim locData(1 To locNumRows, 1 To locNumCols + 1) As Variant
'Copies the file into an array
If parExcludeCharacter <> "" Then
For i = 1 To locNumRows
For j = 0 To UBound(locLinesList(i), 1)
If Left(locLinesList(i)(j), 1) = parExcludeCharacter Then
If Right(locLinesList(i)(j), 1) = parExcludeCharacter Then
locLinesList(i)(j) = Mid(locLinesList(i)(j), 2, Len(locLinesList(i)(j)) - 2) 'If locTempArray = "", Mid returns ""
Else
locLinesList(i)(j) = Right(locLinesList(i)(j), Len(locLinesList(i)(j)) - 1)
End If
ElseIf Right(locLinesList(i)(j), 1) = parExcludeCharacter Then
locLinesList(i)(j) = Left(locLinesList(i)(j), Len(locLinesList(i)(j)) - 1)
End If
locData(i, j + 1) = locLinesList(i)(j)
Next j
Next i
Else
For i = 1 To locNumRows
For j = 0 To UBound(locLinesList(i), 1)
locData(i, j + 1) = locLinesList(i)(j)
Next j
Next i
End If
getDataFromFile = locData
Exit Function
error_open_file: 'returns empty variant
unhandled_error: 'returns empty variant
End Function

Answer 3

This may help you, also it depends how your CSV file is formated.
Open your excel sheet & go to menu Data > Import External Data > Import Data.
Choose your CSV file.
Original data type: choose Fixed width, then Next.
It will autmaticall delimit your columns. then, you may check the splitted columns in Data preview panel.
Then Finish & see.
Note: you may also go with Delimited as Original data type.
In that case, you need to key-in your delimiting character.
HTH!

Answer 4

Have you tried the import text function.

Answer 5

I have the same issue, I'm not able to open a CSV file in Excel. I've found a solution that worked for me in this question Opening a file in excel via Workbooks.OpenText
That question helped me to figure out a code that works for me. The code looks more or less like this:
Private Sub OpenCSVFile(filename as String)
Dim datasourceFilename As String
Dim currentPath As String
datasourceFilename = "\" & filename & ".csv"
currentPath = ActiveWorkbook.Path
Workbooks.OpenText Filename:=currentPath & datasourceFilename, _
Origin:=xlWindows, _
StartRow:=1, _
DataType:=xlDelimited, _
TextQualifier:=xlDoubleQuote, _
ConsecutiveDelimiter:=False, _
Tab:=False, _
Semicolon:=False, _
Comma:=True, _
Space:=False, _
Other:=False, _
FieldInfo:=Array(Array(1, 1), Array(2, 1)), _
DecimalSeparator:=".", _
ThousandsSeparator:=",", _
TrailingMinusNumbers:=True
End Sub
At least, it helped me to know about lots of parameters I can use with Workbooks.OpenText method.

Answer 6

Sometimes all the solutions with Workbooks.open is not working no matter how many parameters are set.
For me, the fastest solution was to change the List separator in Region & language settings.
Region window / Additional settings... / List separator.
If csv is not opening in proper way You probly have set ',' as a list separator. Just change it to ';' and everything is solved.
Just the easiest way when "everything is against You" :P

Answer 7

I just solve (my) problem by setting the Local parameter in the Workbook.Open method like this:
xtrct_wb = Workbooks.Open(filePath, Local:=True)
Every information is then in its proper column. Hope it works for you as well.