Question

reading in a .csv file

Asked by: Ulysses

Good day everyone,

Curious about something.

I am trying to read in a .csv file that has aprox 20 columns if you open it up in excel.
Now my question is this.

When I read it in, first off the columns have headers. Does that go into the first variable?
Second of all not all columns have variables. Does a csv file read an empty column or does it just skip over it?

For some reason i'm having a very hard time reading this .csv file.

Should i just ask the user to save it to excel versus csv?

is there an easier way to do this if there are null value columns?

cheers,
ulysses

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Asked On
2003-10-20 at 20:30:29ID20772717
Topic

Visual Basic Programming

Participating Experts
4
Points
100
Comments
6

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Answers

 

by: pg_indiaPosted on 2003-10-20 at 21:23:18ID: 9588256

I suggest you to save it to excel and then in vb using workbook and worksheet object you can read the excel file. For this you need to add a reference of Excel in your application. or try using createobject method for that...

If you need the code just let me know...

 

by: daffyduck14milPosted on 2003-10-21 at 06:23:24ID: 9590501

Hi there,

The following code will parse your .CSV data and return a neat array. Null values are no problem, since the array position will hold no value (len(x) = 0). That is easy to test for, and should pose no problem.

The code to parse a line of .CSV data is the following. It parses the record you put and returns a array with data.

'---------------------------------------------------------------------------------------
' Procedure : ParseCSV
' DateTime  : 29-09-2003 10:35
' Author    : Donald Lessau (VBSpeed.com)
' Purpose   : Parses a CSV input string
' Inputs    : sExpression: string with the CSV text.
'             asValues(): Array with the values.
' Depends   : None
' Revisions :
'
'---------------------------------------------------------------------------------------

Public Function ParseCSV(ByRef sExpression As String, ByRef asValues() As String) As Long
  Const lAscSpace     As Long = 32   ' Asc(" ")
  Const lAscQuote     As Long = 34   ' Asc("""")
  Const lAscSeparator As Long = 44   ' Asc(","), comma
 
  Const lValueNone    As Long = 0 ' states of the parser
  Const lValuePlain   As Long = 1
  Const lValueQuoted  As Long = 2
 
  ' BUFFERREDIM is ideally exactly the number of values in Expression (minus 1)
  ' so: if you know what to expect, fine-tune here
  Const BUFFERREDIM   As Long = 64
  Dim ubValues        As Long
  Dim cntValues       As Long
 
  Dim abExpression() As Byte
  Dim lCharCode As Long
  Dim posStart As Long
  Dim posEnd As Long
  Dim cntTrim As Long
  Dim lState As Long
  Dim i As Long
' ----------------------
  On Error GoTo PROC_ERR

  If LenB(sExpression) > 0 Then
   
    abExpression = sExpression         ' to byte array
    ubValues = -1 + BUFFERREDIM
    ReDim Preserve asValues(ubValues) ' init array (Preserve is faster)
   
    For i = 0 To UBound(abExpression) Step 2
     
      ' 1. unicode char has 16 bits, but 32 bit Longs process faster
      ' 2. add lower and upper byte: ignoring the upper byte can lead to misinterpretations
      lCharCode = abExpression(i) Or (&H100 * abExpression(i + 1))
     
      Select Case lCharCode
     
      Case lAscSpace
        If lState = lValuePlain Then
          ' at non-quoted value: trim 2 unicode bytes for each space
          cntTrim = cntTrim + 2
        End If
     
      Case lAscSeparator
        If lState = lValueNone Then
          ' ends zero-length value
          If cntValues > ubValues Then
            ubValues = ubValues + BUFFERREDIM
            ReDim Preserve asValues(ubValues)
          End If
          asValues(cntValues) = ""
          cntValues = cntValues + 1
          posStart = i + 2
        ElseIf lState = lValuePlain Then
          ' ends non-quoted value
          lState = lValueNone
          posEnd = i - cntTrim
          If cntValues > ubValues Then
            ubValues = ubValues + BUFFERREDIM
            ReDim Preserve asValues(ubValues)
          End If
          asValues(cntValues) = MidB$(sExpression, posStart + 1, posEnd - posStart)
          cntValues = cntValues + 1
          posStart = i + 2
          cntTrim = 0
        End If
     
      Case lAscQuote
        If lState = lValueNone Then
          ' starts quoted value
          lState = lValueQuoted
          ' trims the opening quote
          posStart = i + 2
        ElseIf lState = lValueQuoted Then
          ' ends quoted value, or is a quote within
          lState = lValuePlain
          ' trims the closing quote
          cntTrim = 2
        End If
     
      Case Else
        If lState = lValueNone Then
          ' starts non-quoted value
          lState = lValuePlain
          posStart = i
        End If
        ' reset trimming
        cntTrim = 0
     
      End Select
   
    Next
   
    ' remainder
    posEnd = i - cntTrim
    If cntValues <> ubValues Then
      ReDim Preserve asValues(cntValues)
    End If
    asValues(cntValues) = MidB$(sExpression, posStart + 1, posEnd - posStart)
    ParseCSV = cntValues + 1
  Else
    ' (Expression = "")
    ' return single-element array containing a zero-length string
    ReDim asValues(0)
    ParseCSV = 1
  End If

PROC_ERR:
  Select Case Err.Number
  Case Is <> 0
    MsgBox "Error " & Err.Number & " (" & Err.Description & ") in procedure ParseCSV of Module modGeneral"
    Err.Clear
  End Select
End Function

Now, to use the function, this is a piece of code that does the job.

Public Sub cmdRead_Click()
  Dim hFile as long ' File handle
  Dim sRecord as string ' Variable to hold a line from the file.
  Dim asRecord() as string ' Array of strings to hold the record data.
' -----------------------------
  On Error GoTo PROC_ERR

  hFile = FreeFile
  Open "C:\Temp\Input.Csv" For Input As hFile

  ' Read in first record.
  Line Input #hFile, sRecord

  ParseCSV sRecord, asRecord
 
  If (asRecord(0) = "Identity") Then
    ' Header seems ok, read in the rest.
    ' Loop through the file and read in one line then proces that line and
    ' do your stuff to it.
  Else
    MsgBox "Format of input file seems invalid.", vbOkOnly+vbInformation
  End If

PROC_EXIT:
  If (hFile <> 0) Then
    ' Close the file.
    Close hFile
    hFile = 0
  End if

PROC_ERR:
  Select Case Err.Number
  Case Is <> 0
    MsgBox Err.Description, vbOkOnly + vbInformation, Err.Number
    Err.Clear
    Resume PROC_EXIT
  End Select
End Sub

The code sample above is crude and not error proof, so care is advised here. Only to give you a general idea of how to work the magic. The ParseCSV function is about the fastest around, and is error free.

Grtz.&copy;

D.

 

by: EddiebabyPosted on 2003-10-21 at 10:25:22ID: 9592247

Hi pg,

Daffy Duck's code looks great, but there is a simpler and more understandable way to handle the problem. It may not be as quick, but it uses commands built into VB6 to help you handle the problem.

In addition, this code doesn't make use of excel which makes it much more efficient - talking to excel through VB can be very slow!

Anyway, the key is to use the split() function. As you know that your headers will be in the first row of the file you can cope with this easily.

I'll chuck it all into a FormLoad so you can see it working.

Private Sub Form_Load()
    '/// Declare our variables
    Dim intHandle As Integer
    Dim strTmp As String
    Dim isFirstLine As Boolean
    Dim strHeaders() As String
    Dim strLines() As String
    Dim intLoop As Integer
    isFirstLine = True

   '/// Set a flag so we know which is our first line
    intHandle = FreeFile

   '/// Open our text file up. You could also use streams, but to make it easier here I'm using
   '/// the old fashioned way of opening up a file.
    Open "c:\<put_your_filename_here>.csv" For Input As intHandle


   '/// Loop until we hit the end of the file.
    While Not EOF(intHandle)

        '/// Get input from the file
        Line Input #intHandle, strTmp

        '/// If its our first line, then it must be the header names.  Using the split command, we put
        '/// them into an array called strHeaders()
        If isFirstLine Then
            strHeaders() = Split(strTmp, ",")
            isFirstLine = False
       Else

           '/// Not headers, so must be data - we put data into our array strLines()
           strLines() = Split(strTmp, ",")

           '/// For ease, I'm going to dump our data out to the immediate window together with the
           '/// field (column) names.
           For intLoop = LBound(strLines()) To UBound(strLines())
                If intLoop <= UBound(strHeaders) Then
                  Debug.Print strHeaders(intLoop), strLines(intLoop)
               Else
                  '/// we seem to have more fields (columns) than headers. Just put Unknown field.
                  '/// we usually get this if too many commas have been used in the data
                  Debug.Print "Unknown Field", strLines(intLoop)
               End If
           Next
       End If
     
      '/// split our output out so that you can see the individual records (rows)

       Debug.Print "/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\"
    Wend
End Sub

This should now work for you. The only problem I know with it really is where the data is packed with too many commas (e.g 1,000,000 would confuse it, better if written to the file as 1000000) but I'm sure you could work a way around that. The easiest way is to have the file saved as tab delimited, much more easier to cope with!!

Hope this helps you.

Eddie

 

by: daffyduck14milPosted on 2003-10-21 at 22:48:35ID: 9596609

Eddie,

Indeed, I concur with your observation the code is not easy to follow. But the problem with using Split is it will cause additional problems if your data is not "clean". As you yourself have said, putting additional comma's in either value objects (your "1,000,000" for example) or string objects ("13th street, corner of whatever" for example)  will yield additional array positions. Hence the use of split is somewhat ackward and not a constant you can rely on.

The function ParseCSV takes into account such "polution" of the data, if and only if there is a value delimiter (usually it's a double quote = " ). I just wanted to point this out, now it is up to Ulysses to decide what fits him best. It's his question. Always explore your options.

Grtz.&copy;

D.

 

by: planoczPosted on 2004-01-13 at 05:24:52ID: 10103669

No comment has been added lately, so it's time to clean up this TA.
I will leave a recommendation in the Cleanup topic area that this question is:
No response from Ulysses from 10/21/2003 comment
Award points to daffyduck14mil is recommend.
Please leave any comments here within the next seven days.
PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER!

planocz
EE Cleanup Volunteer

20120131-EE-VQP-002

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