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The string is like POWER STAR PAVAN , write a script for
this how to findout the number of "A" in the string ?
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Answer / uday
str="POWER STAR PAVAN"
counter=0
For i=1 to len(str)
strChar=mid(str,i,1)
If strChar="A" Then
counter=counter+1
End If
Next
msgbox counter
| Is This Answer Correct ? | 17 Yes | 0 No |
Dim oArray
Dim ochr
oStr="POWER STAR PAVAN"
ochr="A"
oArray=split(oStr,ochr)
print ubound(oArray)
| Is This Answer Correct ? | 6 Yes | 0 No |
Answer / ram
str="POWER STAR PAVAN"
len1=len(str)
str=replace(str,"a", "")
len2=len(str)
a=len1-len2
msgbox a
| Is This Answer Correct ? | 6 Yes | 5 No |
Answer / goms
Val="POWER STAR PAVAN"
If instr(Val,"A")>0 then
Msgbox "Pass"
Else
Msgbox "Fail"
End IF
Note: If instr function returns the position of the value.
If it is greater than 0, the letter is in the particular
variable
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / madhulika
function VerifyLetters(S_String,S_Pattern,MatchStr)
Set OReg = new RegExp
Set S_Pattern = OReg.Pattern
OReg.IgnoreCase = True
OReg.Global = MatchStr
Set Matches = OReg.Test(S_String)
MatchFound = Matches.count
VerifyLetters = MatchFound
End Function
-------------------------------------
val = "Power star pavan"
'Calling function
msgbox VerifyLetters(val,"A",True)
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / seru
str="POWER STAR PAVAN"
Number_of_A=Len(str)-Len(Replace(str,"A",""))
msgbox Number_of_A
Answer is 3
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / ravi
str="POWER STAR PAVAN"
len1=len(str)
msgbox len1
str=replace(str,"A", "")
len2=len(str)
msgbox len2
a=len1-len2
msgbox a
| Is This Answer Correct ? | 1 Yes | 3 No |
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