Find and delete a variable number of spaces following paragraph character

I receive some documents every week that need reformatting. They come with a mixture of non-breaking spaces and normal spaces at the beginning of each paragraph. I could use a simple Find & Replace (optionally embedded in a macro) to find the spurious characters and delete them but their number and distribution ae not the same week to week.

What I need is a few lines of code that find a paragraph mark and then somehow check each subsequent character and if it's a space delete it, continuing until a character in the range [A-Z] is reached. I will pre-process the docs to change all non-breaking spaces into normal spaces.

I'm working in Word 2003.

Thanks in advance.
Dave SwarthoutAsked:
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Rgonzo1971Commented:
Hi,

pls try (using wildcards)
in Find what:
^013(^0032)@

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in Replace with
^013

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Regards
Shaun VermaakTechnical SpecialistCommented:
Have you tried an "Advanced Search" with
^w^v

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Rgonzo1971Commented:
or you could try (without wildcards)
Find what
^013^w

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replace with
^p

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so you do not have to change the non-breaking spaces in "normal" spaces
Rgonzo1971Commented:
As a macro
Sub Macro()
Set rng = ActiveDocument.Range
With rng.Find
    .ClearFormatting
    .Replacement.ClearFormatting
    .Text = "^013^w"
    .Replacement.Text = "^p"
    .Execute Replace:=wdReplaceAll
End With
End Sub

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Dave SwarthoutAuthor Commented:
Thanks much. That macro worked wonderfully.
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