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how to calculate no. of repeating characters in a a
string..please give me the code
- 14 Answers
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Answer / rajeswari
correct code is
charcount=Len(mystring)-Len(replace(Mystring,"s",""))
this one is correct
| Is This Answer Correct ? | 12 Yes | 1 No |
Answer / vijay kumar chourasiya
Dim Str, ChkDup,i,Cnt
Str="aaabc"
For i = 1 to Len(Str) step 1
ChkDup=mid(Str,i,1)
Cnt = split(Str,ChkDup,-1)
result= Ubound(Cnt)
result1=result1&result
Next
MsgBox result1
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / arunkumar vikram r k
Hi Amol, a small change in your script and the repeated characters can be printed or counted
Option Explicit
Dim getChar,strLen,ascChar(),CharCount,i,RepChar,getRes
getChar="VBSsccrript"
strLen=Len(getChar)
ReDim ascChar(strLen)
For i=1 to strLen
ascChar(i)=Mid(getChar,i,1)
Next
For i=1 to strLen
getRes=Split(getChar,ascChar(i),-1,1)
If ubound(getRes)=2 Then
RepChar=RepChar&" "&ascChar(i)
End If
Next
msgbox RepChar
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / sudhanshu
Dim Str, ChkDup,i,Cnt
Str = Inputbox("Enter the String Name")
For i = 1 to Len(Str) step 1
ChkDup=mid(Str,i,1)
Cnt = split(Str,ChkDup,-1)
msgbox Ubound(Cnt)
Next
| Is This Answer Correct ? | 4 Yes | 3 No |
Answer / lakshmi
Hi Rajeswari,
could u please tell us the total no.of repeating charcters
in a given string. it is not specific to one character.
For ex i have a string like "abcdabcefghefg".
so here the total no.of repeating char's are 6.
so for this sring output should be 6.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / kajal
Function RepeatCharCountInString
RepStr = "Automation Using Test Complete"
Msgbox(RepStr)
RepStr = Replace(RepStr," ","")
RepStr = LCase(RepStr)
RepStr1 = RepStr
For i=1 to Len(RepStr)
RepStrChar = mid(RepStr1,i,1)
Count=0
For j=1 to Len(RepStr1)
Char = mid(RepStr1,j,1)
If(Char=RepStrChar) Then
Count = Count +1
End If
Next
If (Count = 0) Then
Exit For
End If
Msgbox(RepStrChar &" Count is " & Count)
RepStr1 = Replace(RepStr1,RepStrChar,"")
RepStr1 = RepStrChar+ RepStr1
Next
End Function
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vinod
str = "abdabcefghefg"
J=0
K=1
z=""
For i=1 to len(str)
x= mid(str,i,1)
If InStr (1, z,x,1) Then
Else
z=z&x
'msgbox z
End If
Next
For i=1 to len(z)
x= mid(z,i,1)
y = split (str,x)
If Ubound (y)>1 Then
varCount=varCount+1
End If
Next
msgbox varCount
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / amol ghule
Option Explicit
Dim getChar,strLen,ascChar(),CharCount,i,UniqChar,getRes
getChar="VBSsccrript"
strLen=Len(getChar)
ReDim ascChar(strLen)
For i=0 to strLen-1
ascChar(i)=Mid(getChar,i+1,1)
Next
For i=0 to strLen-1
getRes=Split(getChar,ascChar(i),-1,1)
If ubound(getRes)=1 Then
UniqChar=UniqChar&" "&ascChar(i)
End If
Next
msgbox UniqChar
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / amol ghule
' A shorter version of code
Option Explicit
Dim getChar,strLen,ascChar(),CharCount,i,UniqChar,getRes
getChar="TestmyKnowledge1221"
strLen=Len(getChar)
ReDim ascChar(strLen)
For i=0 to strLen-1
ascChar(i)=Mid(getChar,i+1,1)
getRes=Split(getChar,ascChar(i),-1,1)
If ubound(getRes)=1 Then
UniqChar=UniqChar&" "&ascChar(i)
End If
Next
msgbox UniqChar
| Is This Answer Correct ? | 0 Yes | 1 No |
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