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