Okay I can find 1 decimal but how do I find two?

General
1

So I poached some code from another program I have and I am trying to figure out how to get it to see two decimals in a string.
Example: 10.2.1
Example: 0.12.2
Example: 0.0.18

Here is the base code I have to work with… I can find the first decimal… but haven’t figured out how to find the second one.

' ---- Variable Defined ----
Var rsData As RowSet = modSel.mthGeneralSelect( "tblChgLog", "DESC" )
Var sIter As String = rsData.Column( "chgIter" ).StringValue
Var iMaj, iMin, iWee, i As Integer
Var strItem As String

' ---- Code In Action ----
i = sIter.IndexOfBytes( "." ) // Find the decimal

If( i = -1 ) Then
  If( sIter = "" ) Then
    strItem = "No Data Found"
  Else
    strItem =  ""
  End If
Else
  strItem = ""
  
End If

rsData.Close

I want to put each number value into a variable for a task, ignoring the decimal.
Example: 10.2.1 iMaj = 10 iMin = 2 iWee = 1
Example: 0.12.2 iMaj = 0 iMin = 12 iWee = 2
Example: 0.0.18 iMaj = 0 iMin = 0 iWee = 18

2 
var Data as string = "10.2.1"
var iMaj as integer = Data.NthField(".", 1).ToInteger
var iMin as integer = Data.NthField(".", 2).ToInteger
var iWee as integer = Data.NthField(".", 3).ToInteger
1 Like
3

I’ll try you method in just a minute… I came up with this crude construct and it works.

' ---- Variable Defined ----
Var rsData As RowSet = modSel.mthGeneralSelect( "tblChgLog", "DESC" )
Var sIter As String = rsData.Column( "chgIter" ).StringValue
Var iMaj, iMin, iWee, i, x As Integer
Var sItem As String

' ---- Code In Action ----
i = sIter.Length

For z As Integer = 0 To i
  If( sIter.Middle( z, 1 ) = "." ) Then
    If( x = 0 ) Then
      iMaj = sItem.ToInteger
      x = 1
      sItem = ""
    Else
      iMin = sItem.ToInteger
      x = i - z - 1
      iWee = sIter.Right( x ).ToInteger
      
      txtIter.Text = iMaj.ToString + " " + iMin.ToString + " " + iWee.ToString
      rsData.Close
      Exit
    End If
  Else
    sItem = sItem + sIter.Middle( z, 1 )
  End If
Next
4

Your title is a bit misleading. WHat you want to do is manipulating string contents. In the Documentation, you have a page for that:

String

5

Yeah… I like your way much better! Less is more they say! Thanks!!!

6

Sure thing! Just make sure you check to make sure your source data is in the right format before you parse it. You can use a regex for that or there are several examples floating around for an InStrCount function to see how many periods are in your source string.

7

Fortunately in this case just the 2. This is a personal program I am working on for myself to track work hours, estimated wages, time with my kids ( I share custody ), expenses and so forth. Essentially I have a program that tracks all the important aspect of my life that I need to keep tabs on in a format that works for me. If you have been through a divorce and child custody stuff… then what I am doing makes sense since I can produce accurate records of time and money flow in and out. And since there is nothing out there that does what I want in a single package… I am writing it! This is my 6th iteration of the program and it’s by far the best.

I have reduced the overall code size by close to 70% with a lot of the things I have learned both from here and from doing. It’s like a light bulb goes on and I see a cleaner way to do what I want. And you just helped me do that again!

1 Like
8

Why not use:

Var  items() as string

parts = sIter.split (".")

Then you have perhaps parts(0), parts(1), and parts(2), which either contain something or don’t exist.

4 Likes
9

I’m going to amend this by saying you may want to add a while loop to make sure there are enough elements, so after splitting:

While ubound(parts) < 2
    Parts.add "0"
Wend

Then you’ll always have 3.

4 Likes
10 
Public Function Get3Parts(dotString As String) As String()
  dotString = dotString + ".."
  Var r() As String = dotString.Split(".")
  Return Array(r(0), r(1), r(2))
End Function

image
image746×325 15.9 KB

image
image749×330 16 KB

Public Function Get3Ints(dotString As String) As Integer()
  dotString = dotString + ".."
  Var r() As String = dotString.Split(".")
  Return Array(r(0).ToInteger, r(1).ToInteger, r(2).ToInteger)
End Function

image
image790×412 15.3 KB

1 Like