replacing_within_range

Bikram · #1 04-22-2022, 04:15 AM

Hello members, I am using the code below posted below to find and replace within range that I have assigned to selection.range but the problem here is it not only replacing within range but beyond the range.

Code:

Sub nnewreplace()
Dim e(), f()
e = Array(" ", " ", Chr(160), Chr(9), "\(([a-z]{1,})\)")
f = Array("", "", "", "", "\1.")
Dim i As Integer
Dim rng As Range
Set rng = Selection.Range
For i = LBound(e) To UBound(e)
With rng.find
    .ClearFormatting
    .MatchWildcards = True
    .Text = e(i)
    .Replacement.Text = f(i)
    .Wrap = wdFindStop
End With
rng.find.Execute Replace:=wdReplaceAll
Next
End Sub

Here all replaced are within range except the last one. I looked by stepping through the code and I found that up to the second last item of the array e(i) the rng was the selection.range but when it passes rng.find.Execute Replace:=wdReplaceAll of last item the rng becomes "" . What can be done to avoid that problem? Any reply would be of great help.

Peterson · #2 04-25-2022, 07:35 AM

Try the code below. By the way, you are searching for spaces twice.

Code:

Sub nnewreplace()
    
    Dim e(), f()
    e = Array(" ", " ", Chr(160), Chr(9), "\(([a-z]{1,})\)")
    f = Array("", "", "", "", "\1.")
    Dim i As Integer
    Dim rng As Range
    Set rng = Selection.Range
    
    For i = LBound(e) To UBound(e)
        With rng.Find
            .ClearFormatting
            .MatchWildcards = True
            .Text = e(i)
            .Replacement.Text = f(i)
            .Wrap = wdFindStop
            .Execute Replace:=wdReplaceAll
        End With
    Next
End Sub

Bikram · #3 04-25-2022, 08:24 PM

Thank you, Peterson, for your suggestion. But the code I posted and the code you replied to seem the same.

Bikram · #4 04-25-2022, 08:27 PM

One thing that I am wondering here is why is it replacing all the items within the selection and only the last item is being replaced outside the selection. Isn't there any way to constrain the replacement within the selection or range?

Peterson · #5 04-25-2022, 10:44 PM

No, it's not the same.

Your code wasn't working because the Execute method was outside of the With statement.

Bikram · #6 04-26-2022, 12:04 AM

Thanks!! Peterson.

Guessed · #7 04-26-2022, 12:32 AM

There is also the potential that the rng is collapsing after the first pass. In my experience, when you use rng.Find the rng collapses to the found instances so a second pass may not include the rng you thought you were looking for.

To avoid this you could redefine the rng before each pass.

Code:

Sub nnewreplace()
    Dim e(), f()
    e = Array(" ", " ", Chr(160), Chr(9), "\(([a-z]{1,})\)")
    f = Array("", "", "", "", "\1.")
    Dim i As Integer, rng As Range
    
    For i = LBound(e) To UBound(e)
        Set rng = Selection.Range
        With rng.Find
            .ClearFormatting
            .MatchWildcards = True
            .Text = e(i)
            .Replacement.Text = f(i)
            .Wrap = wdFindStop
            .Execute Replace:=wdReplaceAll
        End With
    Next
End Sub

Bikram · #8 04-26-2022, 01:22 AM

Thank you very much Andrew, for the reply. I tried to redefine the rng each time as you did but it still replaces the whole document below the selection. The rng collapses mainly on the 4th item after passing .Execute Replace:=wdReplaceAll before this line the rng was the original range. But when the code passes this line on the 4th item I got the rng = "". I couldn't find any explainable reason on the internet so I tried the following code of line temporarily

Code:

    If i = 4 Then
        .Execute Replace:=wdReplaceOne
    Else
        .Execute Replace:=wdReplaceAll
    End If

Guessed · #9 04-26-2022, 02:03 AM

The code I provided looks like it sticks to the selection so I'm not sure why it is failing for you.

Is the first/last character in your selection one of the found items?

Bikram · #10 04-26-2022, 04:46 AM

If the found item is the first character of the selection then wdreplaceall replaces beyond the selection if the found item is other than the first character then the code works fine and only replaces within the selection.