advance find and replace

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Guest

Hi all,

I would like to replace each .0 value with a 0.0. I have tried a find by
using the following wildcard .([0-9]{1}) but it pick up numbers like 15.2
etc. I only need to to add a zero before those numbers that are less than one.
ANy help would be appreciated.
 
Hi all,

I would like to replace each .0 value with a 0.0. I have tried a find by
using the following wildcard .([0-9]{1}) but it pick up numbers like 15.2
etc. I only need to to add a zero before those numbers that are less than one.
ANy help would be appreciated.
Click to expand...

Change the find expression to

([!0-9])(.[0-9])

and use the replace expression

\10\2

The part of the find expression in the first pair of brackets, using
the ! operator to mean NOT, says "any character that is NOT a digit".
The {1} in your expression isn't necessary, since [0-9] already means
"any one digit". So the whole thing says "any character that is not a
digit, followed by a dot and any digit".

The replacement says "the character that matches the first set of
brackets, then a zero, then the characters that match the second set
of brackets".

This replacement will miss an occurrence at the very start of the
document, since there isn't any non-digit character preceding the dot.
That case is easy to fix manually, though.

--
Regards,
Jay Freedman
Microsoft Word MVP
Email cannot be acknowledged; please post all follow-ups to the
newsgroup so all may benefit.
 
Thank you for your help! I tried it and it works great. I did forget to
mention (I apologize) that the values I am trying to replace are in a table
and when I use the expression for that purpose it does not work. Perhaps I am
doing something wrong (I am new to this). If you could follow up with me one
more time I would greatly appreciate it.
Thanks a lot again,
Giusi

Jay Freedman said:
Hi all,

I would like to replace each .0 value with a 0.0. I have tried a find by
using the following wildcard .([0-9]{1}) but it pick up numbers like 15.2
etc. I only need to to add a zero before those numbers that are less than one.
ANy help would be appreciated.
Click to expand...

Change the find expression to

([!0-9])(.[0-9])

and use the replace expression

\10\2

The part of the find expression in the first pair of brackets, using
the ! operator to mean NOT, says "any character that is NOT a digit".
The {1} in your expression isn't necessary, since [0-9] already means
"any one digit". So the whole thing says "any character that is not a
digit, followed by a dot and any digit".

The replacement says "the character that matches the first set of
brackets, then a zero, then the characters that match the second set
of brackets".

This replacement will miss an occurrence at the very start of the
document, since there isn't any non-digit character preceding the dot.
That case is easy to fix manually, though.

--
Regards,
Jay Freedman
Microsoft Word MVP
Email cannot be acknowledged; please post all follow-ups to the
newsgroup so all may benefit.
Click to expand...
 
I think the problem is similiar to what Jay decribes wrt the very start
of the document. In an individual cell there is nothing preceeding the
dot.

Try:
Sub ScratchMacro()
Dim myRng As Range
Dim oCell As Cell
For Each oCell In Selection.Tables(1).Range.Cells
Set myRng = oCell.Range
myRng.End = myRng.End - 1
If myRng.Text = ".0" Then
myRng.Text = "0.0"
End If
Next
End Sub

Jay,

Since I know you will be back and see this, why does the following not
work? I don't understand why the line oCell.Range.End =
oCell.Range.End fails to decrease the range by 1 character:

Sub ScratchMacro2()
Dim oCell As Cell
For Each oCell In Selection.Tables(1).Range.Cells
oCell.Range.End = oCell.Range.End - 1 'This line doesn't cause the
range end value to decrease
If oCell.Range.Text = ".0" Then
oCell.Range.Text = "0.0"
End If
Next
End Sub
 
Oops.

I meant I don't understand why the line oCell.Range.End =
oCell.Range.End - 1 fails to decrease the range by 1 character:
 
Hmm,

I am starting to convince myself that the issue is that a physical
cells range is what it is and can't be modified. This little bit of
code also fails:
Sub ScratchMacro2()
Dim oCell As Cell
For Each oCell In Selection.Tables(1).Range.Cells
oCell.Range.MoveEnd wdCharacter, -1
oCell.Range.Select 'Selects the whole cell :-(
If oCell.Range.Text = ".0" Then
oCell.Range.Text = "0.0"
End If
Next
End Sub
 
Hmm,

I am starting to convince myself that the issue is that a physical
cells range is what it is and can't be modified. This little bit of
code also fails:
Sub ScratchMacro2()
Dim oCell As Cell
For Each oCell In Selection.Tables(1).Range.Cells
oCell.Range.MoveEnd wdCharacter, -1
oCell.Range.Select 'Selects the whole cell :-(
If oCell.Range.Text = ".0" Then
oCell.Range.Text = "0.0"
End If
Next
End Sub
Click to expand...

Hi Greg,

Yes, that's it exactly. You can't modify the .Start and .End of the
ranges of actual objects in the document, such as a cell or a
paragraph or the whole ActiveDocument. What you can do, though, is
assign that range to a Range object you declare, and then tinker with
that:

Sub ScratchMacro3()
Dim oCell As Cell
Dim oRg As Range
For Each oCell In Selection.Tables(1).Range.Cells
Set oRg = oCell.Range
oRg.MoveEnd wdCharacter, -1
If oRg.Text = ".0" Then
oRg.Text = "0.0"
End If
Next
Set oRg = Nothing
End Sub

--
Regards,
Jay Freedman
Microsoft Word MVP
Email cannot be acknowledged; please post all follow-ups to the
newsgroup so all may benefit.
 
Yes Jay, that is pretty much what I did in posting a possible solution
to the OPs question earlier. It was while working that out that I was
perplexed by not being able to simple manipulate the cell range.
Thanks for confirming my thoughts.
 
Thanks for all your help! The macro worked just fine. I have one more
question: can it be generalized to other numbers e.g., .1; .2; .256; .098
etc.? Again, I am new to both macros and wildcards and I don't know how to
change them myself.
Sincerely,
Giusi
 
Giusi,

We have abandoned "wildcards" ;-)

Try:

Sub ScratchMacro()
Dim myRng As Range
Dim oCell As Cell
For Each oCell In Selection.Tables(1).Range.Cells
Set myRng = oCell.Range
myRng.End = myRng.End - 1
If Left(myRng.Text, 1) = "." Then
myRng.InsertBefore "0"
End If
Next
End Sub
 
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