Removing information from a string? VB6

Hello, I have a *.txt file and inside it contains lines with information (each block of information in a new line). This information has a similar pattern:
2014.05.26 11:45:00,67.14975845410586

The next lines have the same structure, the difference is the date value and the number's value.

In the above line I want to remove the comma and all the information in the left. About the number I want the integer part and the first four decimals.
Example: 67.1497

I have tried with Trim, Right, Mid... but I cannot code it right.

You should use InStr() function to get the comma position,
next use Mid$() to retrieve the string at right of comma.
Read help file for how to use this functions.

Code:

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Dim arr() As String
Dim tmp As String

Open App.Path & "\File.txt" For Input As #1
Open App.Path & "\NewFile.txt" For Output As #2
While Not EOF(1)
    Line Input #1, tmp
    arr = Split(tmp, ",")
    Print #2, Round(arr(1), 4)
    DoEvents
Wend
Close #1
Close #2
MsgBox "Done!"

---

Use Instr to get the position of the comma (save it in a variable, e.g. commaPos)
Use Instr to get the position of the period and add 4 to it (save it in a variable, e.g. lastPos)
Subtract commaPos from lastPos to get the Length you want to pull.
Use Mid$ to pull the string starting at commaPos + 1, and the length you calculated.

If you're not sure there will always be four or more digits after the decimal point, you might want to check that lastPos is not greater than the length of the string, and if it is, set it to the length of the string before calculating the length you want to pull.

oops i thought you wanted all of the data - any way the process it the same just amend the output


you can read the lines in as varA and varB

and output varA and varB with separator=space

ie

Code:

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output=varA & " " & fn(varB)

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where the function trims var as required


or work the whole line use replace to replace the comma

Code:

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varB=12.667744
use Format$(varB,"###0.0000") = 12.6677

---

so line above would be

Code:

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output=varA & " " & format$(varB,"###.0000")

---

this handles 3 leading numbers and 4 decimals

so add and remove # appropriately


here to help

remember to rate replies that help and close the thread when done

Why wouldn't you just keep all the decimals and round it?

Dim Val As Double
Dim Line$
Line = "2014.05.26 11:45:00,67.14975845410586"
Val = Fix(CDbl(Split(Line$, ",")(1)) * 10000#) / 10000#

https://support.microsoft.com/en-us/kb/196652

While split will work if that is a fixed position where the comma is then I would just use Mid$() to get either the 7 characters you want or everything after the comma and round it off to 4 places.

If the comma may occur at different positions then I would use Instr() and Mid$() as that would be faster than using Split()

Quote:

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Originally Posted by incidentals

oops i thought you wanted all of the data - any way the process it the same just amend the output

---

Probably off on a tangeant there. In post #1 it was implied this is existing data taken from a file, to be "cleaned." Not data being created by the program.


There could be other issues too. Might be lines with no comma. Might be spaces after the comma to be trimmed. Might be lots of things. All can be handled by tweaking the code:

Attachment 138351

Before


Attachment 138353

After

Code:

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Option Explicit

Private SuppressEvents As Boolean

Private Sub cmdClean_Click()
    Dim Text As String
    Dim OutPos As Long
    Dim InPos As Long
    Dim CommaPos As Long
    Dim EndLinePos As Long
    Dim MoveLen As Long
    Dim ChunkToMove As String

    cmdClean.Enabled = False
    Text = Text1.Text
    OutPos = 1
    InPos = 1
    Do While InPos <= Len(Text)
        CommaPos = InStr(InPos, Text, ",")
        If CommaPos < 1 Then CommaPos = InPos - 1
        EndLinePos = InStr(InPos, Text, vbNewLine)
        If EndLinePos < 1 Then
            EndLinePos = Len(Text) + 1
        Else
            EndLinePos = EndLinePos + 2
        End If
        If CommaPos < EndLinePos Then InPos = CommaPos + 1
        MoveLen = EndLinePos - InPos
        'Remove leading spaces.  If not desired, remove
        'use of LTrim$() here:
        ChunkToMove = LTrim$(Mid$(Text, InPos, MoveLen))
        Mid$(Text, OutPos) = ChunkToMove
        InPos = InPos + MoveLen
        OutPos = OutPos + Len(ChunkToMove)
    Loop
    SuppressEvents = True
    Text1.Text = Left$(Text, OutPos - 1)
    SuppressEvents = False
End Sub

Private Sub Text1_Change()
    If SuppressEvents Then Exit Sub
    cmdClean.Enabled = True
End Sub

---

Zip, zam, zowie. Much quicker than "split-de-splits" or awful RegEx techniques, though not important unless large masses of data need to be cleaned.

Also, showing leading space removal:

Attachment 138357

Add a comma


Attachment 138359

After

op does not need to think about errant lines he states all lines are the same structure

so read a,b

will resolve the number with the decimal is b

so format$(b,"###.0000") would be fine

how the op uses these facts is upto them


here to help

You might want to re-read the question. It isn't about output but modifying input.

the process of modifying a thing that comes in is to output the modification

whether the change is made to a line in a file or a placeholder in memory the approach is the same

the op needs to know that the line in the file can be read in as a and b where b is after the comma

and that b can be modified using the format$ function...

Now tell me how do you change the contents of a file so that it reflects the needs of the op without "outputing"/modifying something

where are they putting/keeping the changed value/shape or B

here to wonder if you are reading what i am writting?


its about the OP learning some thing new

(we should not be fighting each other here)

here to be as helpful as possible

what follows is an example of what the op could do using the technique i have suggested

there is enough here to help the OP understand how data can be read and manipulated in vb6

Code:

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Dim strEmpFileName As String
Dim strBackSlash As String
Dim intEmpFileNbr As Integer
 
Dim strEmpName As String
Dim intDeptNbr As Integer
Dim strJobTitle As String
Dim dtmHireDate As Date
Dim sngHrlyRate As Single
 
strBackSlash = IIf (Right$(App.Path, 1) = "\", "", "\")
strEmpFileName = App.Path & strBackSlash & "EMPLOYEE.DAT"
intEmpFileNbr = FreeFile
 
Open strEmpFileName For Input As #intEmpFileNbr
 
Do Until EOF(intEmpFileNbr)

Input #intEmpFileNbr, strEmpName, intDeptNbr, strJobTitle, dtmHireDate, sngHrlyRate
Print strEmpName; _

Tab(25); Format$(intDeptNbr, "@@@@"); _

Tab(35); strJobTitle; _

Tab(55); Format$(dtmHireDate, "mm/dd/yyyy"); _

Tab(70); Format$(Format$(sngHrlyRate, "Standard"), "@@@@@@@")

Loop

---

here to help