VB Script Quiz: How will you retrieve a Number from a dynamic String with dynamic length?

by | Last Updated On Oct 3, 2016 | Test Yourself on QTP | 29 comments

uft-extract-number

In test automation scripting you have to deal a lot with strings and numbers and should know how to extract or retrieve one from the other. Sometime back, there was an interesting question posted on a forum. I am not posting the exact question here but its variation for more clarity.

Given a dynamic string, for example 654ABC or 5432FGHJS or 9GFHSK , how will you extract all the numbers that appear in the front part of the string?

Your code should be adaptable enough to take any string and output the numerical digits appearing in-front part of the string.

The answers we are looking from above strings are 654, 5432 and 9.

Please post your code in the comments section below.

Answer

We are really overwhelmed by the response of UFT community towards the quiz. There were some excellent answers there.
There is no right or wrong solution here. At times we may know all the methods to solve a problem but we tend to use the one which strikes us first.

Checking how others would approach and solve the same problem is the best way to sharpen your scripting skills.

While there can be multiple ways to arrive at the solution, my preferred way in this case would be to use RegExp object to find the digits quickly. Using For-Loop is a performance intensive method.

Dim s : s = "654ABC"

Set re = New RegExp 'Create Regular expression object
re.Pattern = "^\d+" 'Matches if a string starts with a digit

Set matches = re.Execute(s) 'Collection

If matches.Count <> 0 Then
msgbox matches(0).Value 'Output the first value if it exists
else
msgbox "The input string " & s & " doesn't start with a digit"
End If

You Should Read These 3 Related Articles Too!

29 Comments

  1. Ankur Jain on Oct 3, 2016 at 13:44

    Thank you all for the excellent answers. It certainly gives a perspective on how different people think differently while arriving at the same answers.

    We have updated our answer in the post above.

    Reply
  2. Raja on Oct 1, 2016 at 07:54 
    str_input=Array("654ABC","5432FGHJS","9GFHSK","ABC10","ABCD","123ABC986")
str_FailMsg = "No Numbers appears in the first part of the given string" 
For intLP=0 to ubound(str_input)
	strOutput=""
	For intInLp=1 to len(str_input(intLp))
		If asc(mid(str_input(intLp),intInLp,1))>=48 and asc(mid(str_input(intLp),intInLp,1))<=57 Then
			If isnumeric(mid(str_input(intLp),intInLp,1)) Then
				strOutput = strOutput & mid(str_input(intLp),intInLp,1)
			End iF	
		Else
			Exit For
		End If
	Next
	IF strOutput="" Then
		print str_FailMsg
	Else
		print strOutput
	End IF
Next
    Reply
  3. Chinnikrishna Darapureddy on Sep 30, 2016 at 22:45 
    x="A12345BC"
set regExp=new RegExp
regExp.Pattern="[0-9]"
regExp.Global=True
regExp.IgnoreCase=True
Set Matches = regExp.Execute(x)   ' Execute search.
	For Each Match in Matches   ' Iterate Matches collection.
		int_Val = Match.Value
		msgbox int_Val 
	Next

'it gives 12345 as popup
'if we give string as x="12ABC1256"
'then it gives two times popup with 12, 1256
    Reply
  4. Mohammed Jabbar on Sep 30, 2016 at 22:34 
    myString = "654ABC"
myStringLength = Len(myString)
'msgbox myStringLength


myString = Left(myString, 3)
msgbox myString
    Reply
  5. Ram.S on Sep 30, 2016 at 21:11 
    ' -------------------------------------------------------------------------------
' Function : To Extract numbers from a string at any position & not just starting
' Date: 9/30/2016
' Author: Ram S
' Project : LearnQTP Quiz
' Revision: 1.0
' -------------------------------------------------------------------------------
Option Explicit

Dim sstring, objRegEx, valMatch, inString, oString

Set objRegEx = CreateObject("VBScript.RegExp")

objRegEx.Global = True

objRegEx.Pattern = "[^A-Za-z]"

sstring = "5432FGHJS"

Set inString = objRegEx.Execute(sstring)  

If inString.Count > 0 Then

    For Each valMatch in inString

        oString = oString & valMatch.Value

    Next

    Wscript.echo oString

End If

' --- Output = 5432
    Reply
  6. Tejhaswini on Sep 30, 2016 at 18:57 
    'Function to fetch the numeric data from a dynamic string

Function FunRetrieveNumericDataFromString(vString)
vNumericData = ""
vStringLen = Len(vString )
for i = 1 to vStringLen 
vNumeric = Mid(vString ,i,1)
If IsNumeric(vNumeric ) then
    vNumericData = vNumericData&vNumeric 
End If
FunRetrieveNumericDataFromString= vNumericData 
Next
End Function

'Ex:
vString = "654ABC"
'By calling the below function we can get the Numeric data from a dynamic string.
vRetrieveNumericData = FunRetrieveNumericDataFromString(vString)
    Reply
  7. joe Daniel Lawrence on Sep 30, 2016 at 17:39 
    s = "9GFHSK"
n = Len(s)
result = ""

for i=1 to n
	c = Mid(s, i, 1)
	if asc(c) >= 65 Then
		i = n+1
	else
		result = result + c
	end if


next

msgbox result
    Reply
    • Venkateswara Rao Saladi on Sep 30, 2016 at 18:10

      Answer : 9

      Reply
    • Michael Bowen on Sep 30, 2016 at 21:50 
      Option Explicit

Sub FindNumberInString()
    'Variable declaration
    Dim strValue As String                                                          'Input String
    Dim strLen As Integer                                                           'Holds length of input string
    Dim OutString As String                                                         'Holds Output String
    Dim i As Integer                                                                        'Index variable
    
    strValue = Worksheets("Sheet1").Range("B1").Value   'Read value into variable
    strLen = Len(strValue)                                                              'Find length of entire string
    OutString = ""                                                                              'Zero OutString
    For i = 0 To strLen - 1                                                                 'Progress through each item
        Dim strItem As String                                                           'Holds individual values as they are removed
        strItem = Mid(strValue, i + 1, 1)                                         'removes item
       If IsNumeric(strItem) Then                                                   'Test to see if it is a number
            OutString = OutString + strItem                                     ' Concatenates to OutString
        End If
    Next i
    Worksheets("Sheet1").Range("B2").Value = OutString  'Write OutString
End Sub
      Reply
  8. Subh on Sep 30, 2016 at 17:31 
    str1=InputBox("Enter the string")
strLength=len(str1)
For X = 1 To strLength  
    If IsNumeric(Mid(str1, X, 1)) = True Then  
        nNum = nNum & Mid(str1, X, 1)  
else 

exit for
    End If  
Next   
MsgBox nNum
    Reply
  9. Hung Cao on Sep 30, 2016 at 17:29

    There are two ways to do it…. It is depend what is your requirement and expect the result.

    1. Non digit is “/D” so you can use the replace function to replace “/D” to “”. The drawback on this is following 123XX2V would return 1232. XX12ER3 would return 123.

    2. Write a function except the string for input and return the number. Use the for loop to check every character. ignore when you see the char. 123XX2V would return 123. XX12ER3 would return 12.

    Reply
  10. Phamo on Sep 30, 2016 at 16:49 
    Public Function ExtractNumericFromString(text)
	Set regEx = New RegExp
	regEx.Pattern = ".*\d+.*"
	regEx.IgnoreCase = True
	regEx.Global = True

	If regEx.Test(text)Then
		regEx.Pattern = "[^0-9]"
		results = regEx.Replace(text,"")
	Else
		resuts = ""
	End If
	ExtractNumericFromString = results
End Function

Wscript.Echo "654ABC: " & ExtractNumericFromString("654ABC")
Wscript.Echo "5432FGHJS: " & ExtractNumericFromString("5432FGHJS")
Wscript.Echo "9GFHSK: " & ExtractNumericFromString("9GFHSK")
    Reply
  11. ishan on Sep 30, 2016 at 16:40 
    Sub test()
    Str1 = "3test4test"
    strlen = Len(Str1)
    
    
    For i = 1 To strlen
        
        char1 = Mid(Str1, i, 1)
        
        If IsNumeric(char1) Then
        
            num_val = num_val & char1
        End If
    Next
    
    
    MsgBox num_val
 End Sub
    Reply
  12. Chandan on Sep 30, 2016 at 16:15 
    str = "5432FGHJS" ''Enter the required String here
num_part = ""
For i=1 to len(str)
	temp = right(left(str,i),1)
	If IsNumeric(temp) = True Then
		num_part = num_part & temp
	Else
		Exit For
	End If
next
MsgBox "The Numeric Part in the Front is: " & num_part
    Reply
  13. Deepak on Sep 30, 2016 at 16:07 
    On error resume next
str = "654ABC"
numstr =""
For i = 1 To len (str) Step 1
	
	strnum = cint (mid (str, i, 1))
If err.number = 0 Then
	numstr = numstr & strnum 
	Print "Numbers from the string '" & str &"' is :" & numstr
Else
	'msgbox numstr
	Exit For
End If
Next
    Reply
  14. srinivasu ch on Sep 30, 2016 at 16:01 
    x="A12345BC"
set regExp=new RegExp
regExp.Pattern="[\d]+"
regExp.Global=True
regExp.IgnoreCase=True
Set Matches = regExp.Execute(x)   ' Execute search.
	For Each Match in Matches   ' Iterate Matches collection.
		int_Val = Match.Value
		msgbox int_Val 
	Next

'it gives 12345 as popup
'if we give string as x="12ABC1256"
'then it gives two times popup with 12, 1256
    Reply
  15. Manya on Sep 30, 2016 at 15:31 
    Dim myStr, myDigits

myStr = "7898tfgy"

nLength = Len(myStr)

For Iterator = 1 To nLength

	nDigit = Mid(myStr,Iterator,1)
	If IsNumeric(nDigit) Then
		myDigits = myDigits&nDigit
	End If
	
Next
msgbox myDigits
    Reply
  16. Raghavendra Rao Unnam on Sep 30, 2016 at 15:29 
    Dim RegEx : Set RegEx = New RegExp
RegEx.Pattern = "\d+"
RegEx.Global = True
Dim strTemp : strTemp = "654ABC or 5432FGHJS or 9GFHSK"

Set matches = RegEx.Execute(strTemp)

For i = 0 To matches.count - 1
	Reporter.ReportEvent micPass,matches.item(i),matches.item(i)
Next
    Reply
  17. Jagan on Sep 30, 2016 at 15:25 
    'My input String
MyStr = "023Abc236"  

'Performing loop for entire string length
For i = 1 To len(str) 

' Reading one character at a time
	val = mid(str,i,1)

'Comparing each string with ascii value between range 48 to 57 ( which is numbers)
		If asc(val) >=48 and asc(val) <= 57 Then 

'If condition is TRUE Printing the Numbers that appear front of string	
			print "Numbers that appear in the front part of the given string is: " & val  
	
		else 

'If condition is FALSE Printing that NO Numbers appear front of string	
			print "Sorry, no numbers appear in the front part of the given string." 

'Forcing FOR LOOP to STOP and EXIT as there is no objective in going further.			
			Exit for 
	
		End If 
	
Next
    Reply
  18. vikash on Sep 30, 2016 at 15:22 
    Dim intval 'varibale declaration
Dim intval1
dim inval2

intval ="654ABC"  'or 5432FGHJS or 9GFHSK 
for i= 1 to len(intval)
   inval2=mid(intval,i,1)
   if IsNumeric(inval2) then
      counter= counter+inval2
   end if
next
msgbox counter
    Reply
  19. Amit Kumar Singh on Sep 30, 2016 at 15:16 
    sMystring= "5432FGHJS"
numericStr= ""
For i= 0 to len(sMystring)
	sSplittedval= Mid(sMystring, i+1, 1)
	If isNumeric(sSplittedval) Then
		numericStr= numericStr & sSplittedval
	End If
Next
msgbox trim(numericStr)
    Reply
  20. Hari on Sep 30, 2016 at 15:14 
    StrValue = InputBox("Please Enter String")

Set objRgExp = New RegExp
objRgExp.Global = True
objRgExp.Pattern = "[0-9]+"
Set strRes = objRgExp.Execute(StrValue)
For Each res in strRes
	msgbox res.value,0,"String Found"
Next
    Reply
  21. Ravi on Sep 30, 2016 at 15:11 
    'Taking example string as 1234ABC

str1="1234ABC"

leng=len(str1)

For i = 1 To leng
	substr=mid(str1,i,1)
	If isnumeric(substr) Then
		Print substr
	Else
		pos=i
		Exit for
	End If
Next

res=mid(str1,1,pos-1)
msgbox res
    Reply
  22. Rohit Shinde on Sep 30, 2016 at 15:07

    Hi Ankur,

    Please see the below piece of code to find the dynamic legnth number form dynamic string.

    Dim str, strLen, num
str = "654ABC"
strLen = Len(str)
num = ""

For i = 1 To strLen
	If IsNumeric(Mid(str, i, 1)) Then
		num = num & Mid(str, i, 1)
	End If
Next

Print num
    Reply
  23. Lokesh Kumar on Sep 30, 2016 at 15:05 
    str = "786abhikansh567st1239test"
Set regEx = New RegExp 
regEx.pattern="^\d+"
regEx.Global = True
Set obj = regEx.Execute(str)

For i=0 to obj.count-1

	temp = temp & obj.item(i)

Next

msgbox temp
    Reply
  24. Raghavendra Rao Unnam on Sep 30, 2016 at 15:03 
    Dim arrTemp(10)
intArrIndex = 0
strVal = "654ABC or 5432FGHJS or 9GFHSK"

For intIndex = 1 To Len(strVal)
     
       chrVal = Mid(strVal,intIndex,1)
       If IsNumeric(chrVal) Then
       	  strTemp = strTemp & cstr(chrVal)
       	  If Not isNumeric(Mid(strVal,intIndex + 1,1)) Then
       	      arrTemp(intArrIndex) = strTemp
       	      strTemp = ""
       	      intArrIndex = intArrIndex + 1
          End If      	  

       End If
 	
Next

For intIndex = 0 To intArrIndex - 1
msgbox arrTemp(intIndex)
	
Next
    Reply
  25. Jyotikrushna Mohapatra on Sep 30, 2016 at 14:55 
    st="654ABC"
x=len(st)
Temp=""

For i=1 to x
	y=mid(st,i,i)
	If isnumeric(y)="True" Then
    Temp=Temp&y
    Else
		Exit for
	End If

Next
msgbox Temp
    Reply
    • Jyotikrushna Mohapatra on Sep 30, 2016 at 15:06 
      st="12345hgjkgjkhjkh"
x=len(st)

For i=1 to x
	y=mid(st,i,1)
	If isnumeric(y)="True" Then
     Temp = Temp&y
    Else 
		Exit for
	End If

Next
msgbox Temp
      Reply
  26. Harry McKay on Sep 30, 2016 at 14:50 
    Dim test_string, all_numbers, string_length, loop_counter, numbers_only, test_char

test_string = "5432FGHJS"

all_numbers = "0123456789"

string_length = Len(test_string)

For loop_counter = 1 To string_length

	test_char = Mid(test_string, loop_counter, 1)
	
	If InStr(1, all_numbers,test_char , 1) = 0 Then
		
		numbers_only = Left(test_string, (loop_counter - 1))
		
		Exit For
		
	End If
	
Next

reporter.ReportEvent micPass, "original string = " & test_string, "extracted numbers = " & numbers_only
    Reply

Submit a Comment

Your email address will not be published. Required fields are marked *