B4J Question how to find intervals i a string?

danchovasilev · 2024-12-21T16:33:51+0000

in VB my code was
For i = 1 To Len(ArG)
If Mid(ArG, i, 1) = " " Then
BrIntervali = BrIntervali + 1
End If
Next i

Erel · 2024-12-22T06:37:06+0000

B4X:

Dim ArG As String = "in VB my code was"
Dim BrIntervali As Int
For i = 0 To ArG.Length - 1
    If ArG.CharAt(i) = " " Then BrIntervali = BrIntervali + 1
Next
Log(BrIntervali)

emexes · 2024-12-22T07:29:22+0000

Is the next step to divide the string, splitting it at each space?

What if there are two spaces together? Is there an empty (zero length) field between the two spaces?

What about leading and trailing spaces?

Your simplest solution might be Regex.Split:

B4X:

LogStringArray(Regex.Split(" ", "Now is the time"))
LogStringArray(Regex.Split(" ", " for all good men "))
LogStringArray(Regex.Split(" ", "to come to the"))
LogStringArray(Regex.Split(" ", " aid of the party."))

Sub LogStringArray(S() As String)

    Log("Split into " & S.Length & " substrings")
    For I = 0 To S.Length - 1
        Log(I & TAB & """" & S(I) & """")
    Next
 
End Sub

Log output:

Split into 4 substrings
0    "Now"
1    "is"
2    "the"
3    "time"
Split into 5 substrings
0    ""
1    "for"
2    "all"
3    "good"
4    "men"
Split into 4 substrings
0    "to"
1    "come"
2    "to"
3    "the"
Split into 5 substrings
0    ""
1    "aid"
2    "of"
3    "the"
4    "party."

emexes · 2024-12-22T13:32:30+0000

emexes said:
Your simplest solution might be Regex.Split:

danchovasilev · 2024-12-22T13:33:20+0000

danchovasilev said:
in VB my code was
For i = 1 To Len(ArG)
If Mid(ArG, i, 1) = " " Then
BrIntervali = BrIntervali + 1
End If
Next i

First of all, thank you very much for your answers.
I did it like this after searching the forum...

my example:

Sub interval(mstr As String) As Int
    
    Dim pattern As String
    pattern = " "
    Dim Matcher1 As Matcher
    Dim i As Int
    i=0
    Matcher1 = Regex.Matcher(pattern, mstr)
        Do While Matcher1.Find
            i=i+1
            Log("Found: " & i)
            Log("----: " & Matcher1.Match)
        Loop
    Return i   
End Sub

emexes · 2024-12-22T15:00:30+0000

danchovasilev said:

my example:

Sub interval(mstr As String) As Int
   
    Dim pattern As String
    pattern = " "
    Dim Matcher1 As Matcher
    Dim i As Int
    i=0
    Matcher1 = Regex.Matcher(pattern, mstr)
        Do While Matcher1.Find
            i=i+1
            Log("Found: " & i)
            Log("----: " & Matcher1.Match)
        Loop
    Return i  
End Sub

If you just wanted a count then you could do:

B4X:

Sub interval(mstr As String) As Int
    Return Regex.Split(" ", mstr.Trim).Length
End Sub

emexes · 2024-12-22T15:12:02+0000

I usually use Regex.Split because it puts all the fields (substrings) into an array ready for whatever the next processing step is.

The .Trim removes leading and trailing spaces.

You can just as easily use commas or semicolons or tabs as separators.

B4J Question how to find intervals i a string?

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