Find number in a string

[pjeigenn]

Hi, I am looking a function that takes a number from a string.

For example I have:

"The number of players was 86 and..."

Is there any native function of vbScripting that does this?

Thanks

[MVChowdary]

Hello,
Below is the example code to get the number from string.

Code:

l_String="The number of players was 86 and..."
l_num=""
    For i=1 to len(l_String)
            y=mid(l_String,i,1)
                If ( isnumeric(y)=True) Then
                          l_num= l_num & y
                End If
    Next
    msgbox l_num

[Arun Prakash]

Code:

strString="The number of players was 86 and 88 hkjh 880"

    splitString=Split(strString," ")
    intCount= UBound(splitString)
    intCount= Cint(intCount)
    arrayNum=" "
    For i = 0 to intCount
        If IsNumeric(splitString(i)) Then
            arrayNumbers= splitString(i)
            If arrayNumbers<>""  and arrayNum =" "  Then
                arrayNum= arrayNumbers            
            Else
                arrayNum= arrayNum &"_"& arrayNumbers            
            End If
            
        End If
    
    Next
    MsgBox arrayNum

[pjeigenn]

Thanks. Looking the string, it has several numbers, but the number i need is next to the word "código" (code in English), it can be done with regular expressions? Or I have to add some lines to the code below.

I will think the solution, if it works i will post it.

---

Solved

Code:

l_String=Datatable("o_DescRechazo", dtLocalSheet)
l_num=""
    For j=1 to len(l_String)
       y=mid(l_String,j,6)
       If ( y="Código") Then
         For i=j to len(l_String)
           y=mid(l_String,i,1)
           If ( isnumeric(y)=True) Then
              l_num= l_num & y
            End If
          Next
          j=len(l_string)                        
       End If
     Next
  
    msgbox l_num

Thanks to all

[basanth27]

I understand it is solved. Just curious to know, If your text string was so,

txtstrng = "this is codeigo 86 testing numeric"

I would simply do this,

Code:

reqnum = Split(Split(txtstrng,"codeigo")(1),"testing")(0)
msgbox reqnum

Why run unnecessary loops ?

Food for thought mate...

[pjeigenn]

It is an interesting solution... the "código" word is always going to be before the number i need, but i don´t know what is next to that word.
Btw, i don´t completely understand what are you trying to do there, why (1)? why (0)?

It is an interesnting alternative, with better performance...

Although it doesn´t work "The subindex is out of range: '[number: 1]'"

[supputuri]

Hi,
Let me explain the below in detailed.

Code:

txtstrng = "this is codeigo 86 testing numeric"

reqnum = Split(Split(txtstrng,"codeigo")(1),"testing")(0)
Let us look into the inner split first and then the outer split.
When you use split method using "codeigo", then it will breake into 2 strings.
Split(txtstrng,"codeigo")(1):
  String(0) = "this is codeigo"
  String(1)= "86 testing numeric"
So, Split(txtstrng,"codeigo")(1) = "86 testing numeric"
Now here again we are spliting the string with "testing"

  Split(Split(txtstrng,"codeigo")(1),"testing")(0):
  Split("86 testing numeric","testing")
When you use split method using "testing", then it will breake into 2 strings.
string(0)= "86 "
string(1)="testing numeric"

As we need the numeric value. We have to capture the first string i.e. string(0).


Please let me know if you still have any questions.

[ngocvo3103]

Hi there,

You can use this way to get any number(s) from a string

Code:

Public sub s_GetNumberFromString (strText, byref colNum)
    Set ObjRegExp = New RegExp
    ObjRegExp.Pattern = "\d+"
    ObjRegExp.Global = True
    Set colNum = ObjRegExp.Execute(strText)
End Sub

Dim colNum
s_GetNumberFromString strNormal, colNum
For each intNum in colNum
    MsgBox intNum
Next

Ngoc Vo

[cdesserich]

Code:

Dim regEx: Set regEx = New RegExp
regEx.Pattern = "(?:código )(\d+)"
MsgBox regEx.Execute("this is código 86 testing numeric").Item(0).SubMatches.Item(0)

(?: ) is a non-capturing group, so you are assured the digits will end up in the first SubMatch of the SubMatches collection. If you want to break the code down a little further:

Code:

Dim txtstrng: txtstrng = "this is código 86 testing numeric"

Dim regEx: Set regEx = New RegExp
regEx.Pattern = "(?:código )(\d+)"

Dim regExMatches: Set regExMatches = regEx.Execute(txtstrng)

For i = 0 To regExMatches.Count-1
    MsgBox "código match number " & i & ": " & regExMatches.Item(i).SubMatches.Item(0)
Next

This will iterate through each match if there are more than one.

[bfakruddin]

Code:

Dim var
var = inputbox("Enter the string with any numeric value", "Alpha Numeric", "Baba205Fakruddin")

Function Get_No_FromString(var)

    len_var = len(var)

    bln=""
    For i=1 to len_var
    
        cha =mid(var,i,1)
    
        If isnumeric(cha) Then          '1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or  9)
            bln = bln&cha
            print bln
        End If
    
    Next
    
    print "The digits in string"&bln

    Get_No_FromString=bln
End Function


No_in_String = Get_No_FromString(var)
msgbox "The digits available in string"&No_in_String