Answer / udayakiran

This might be the solution If we consider it as array:

ar2 = Array("C","A","P","G","E","M","I","N","I")

arlen=UBound(ar2)-lbound(ar2)+1 'Finding array length

For p= 0 to arlen-1
If strcomp(ar2(p),"G")=0 Then
msgbox "The letter G is found"
Exit For
End If
Next

Answer / tarakeshwar

a="CAPGEMINI"
length = len(a)
flag ="G is not present"
For i = 1 to length
res= mid(a,i,1)
if res = "G" then
flag = "G is present"
End if
Next
msgbox flag

Answer / arun

Please try this

Dim AR2,i
AR2 = Array("C","A","P","G","E","M","I","N","I")

For i=0 to ubound(AR2)
If AR2(i)="G" Then
msgbox "G found in Array"
End If
Next

Answer / gabbar

Dim a
a=array("CAPGEMINI" )
If len(join(a))<>len(Replace(join(a),"G","")) Then
msgbox "Occurance of G in array: "&len(join(a))-len(Replace(join(a),"G",""))
End If

Answer / koteswar

str = "CAPGEMINI"
z= "G "
flag = 0
For i = 1 to len(str)
x = mid(str, i, 1)
If Instr(1,z,x,1) Then
msgbox z
If flag = 0 Then
Reporter.ReportEvent micPass, "The required Letter is
available in the stirng", "True"
else
Reporter.ReportEvent micFail , " The required Letter is
not availabe in the string" , "False"
End If
End IF
Next

Array means It is a collection of similar objects and
enclosed with curly braces. at that time you can use Ubound
for count the no of objects are available in the given array

Srring means single or collection alphanumeric characters
enclosed with quotations. If you want to count the no of
characters in the given string you can use len()

Answer / rajesh

see that is my question...here you also confused, here
CAPGEMINI is a array ,not a string..give a ans please

Answer / ganesh

ok koteshwar can u explain about flag = 0

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