This modification of my modification of Bob's original code seemed to work ok:

Option Explicit
Sub JI2()
Dim pos As Long
Dim CELL As Range
Dim myPositions() As Long
Dim pCtr As Long
Dim FoundOne As Boolean

For Each CELL In Selection
FoundOne = False
pCtr = 0
Do
pos = InStr(1, CELL.Value, "HL", vbBinaryCompare)
If pos = 0 Then
Exit Do
End If
FoundOne = True

pCtr = pCtr + 1
ReDim Preserve myPositions(1 To pCtr)
myPositions(pCtr) = pos
CELL.Value = Left(CELL.Value, pos - 1) & "P" _
& Right(CELL.Value, Len(CELL.Value) - pos - 1)
Loop

If FoundOne = True Then
For pCtr = LBound(myPositions) To UBound(myPositions)
CELL.Characters(myPositions(pCtr), 1).Font.Name = "Wingdings 2"
Next pCtr
End If
Next CELL

End Sub

Shetty wrote:

Thanks to both of you.
Yes Bob. There may be multiple occurances as well as nil occurances. I
tried your code with do loop but it does not work, gives very vague
results. I don't know why it works that way with do loop.

Dave - I have added one If Not IsEmpty(CELL) Then statement to skip
blank cells.
Your code works fine as long as there is atleast one occurance.
If the cell does not contain HL then 1st part to replace is skipped by
If pos = 0 Then
Exit Do
End If
But, 2nd part is still executed. Its a for loop.
For pCtr = LBound(myPositions) To UBound(myPositions)
CELL.Characters(myPositions(pCtr), 1).Font.NAME = "wingding 2"
CELL.Characters(myPositions(pCtr), 1).Font.Bold = True
Next pCtr

There are two things.
1. If 1st cell is with nil occurance, then code breaks on error
highlighting above for loop (subscript out of range).
2. If If any subsequent cell is with nil occurance, then above for loop
changes the font of any other cherecter making it unreadable.

I tried to bypass this for loop with a goto statement.
If pos =0 then goto ncell
...code
ncell:
next cell
end sub
With this, cells with nil occurance are not changed but the replacement
made in cells containing HL are not applied new font.

I hope, this also can be solved or a workaround is possible.
Regards,
Shetty.

--

Dave Peterson