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find and replace shading

Hi all,
I have a document where parts of the text are highlighted or shaded in light yellow. It seems I can't use standard search and replace to replace the highlighted text.

Can anyone help me to find a way to find the shaded parts in the document?
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Shawn
Asked:
Shawn
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1 Solution
 
ShawnAuthor Commented:
this is what I need to find
.BackgroundPatternColor = wdColorYellow
and need to replace any highlighted text with "" (nothing).

The end goal is to count the words that are not highlighted.
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ParanormasticCryptographic EngineerCommented:
You can do that with the search and replace feature, but you need to use the advanced options.

- Ctrl + H
- 'More' button
- 'Format' button
- Highlight (Highlight appears under box for Find what)
- Repeat last 2 steps for Replace With box
- Use the Highligher icon in the main Word window (the pencil with ab, next to character color) and select the desired new color - select 'none' to erase.
- Replace all, or Find Next and Replace as desired if you want to keep some. * note this will change all highlighted items if you replace all, not just any particular color.  (for font/text color, you can actually look for the specific color and be more granular, for backgrounds/highlight you cannot).

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ShawnAuthor Commented:
I have tried this already but no success
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ShawnAuthor Commented:
it is not actually the "highlight" function I am looking for it is the  .BackgroundPatternColor = wdColorLightYellow.

the results can be similar looking but I'm pretty sure BackgroundPatternColor  needs to be searched via  a macro using VB
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Joanne M. OrzechManager, Document Services CenterCommented:
Deleted post.
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ShawnAuthor Commented:
??
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ShawnAuthor Commented:
got it...thanks anyway.

here's the end result...works great

Sub FindHighlightedText()

Selection.HomeKey Unit:=wdStory

With Selection.Find
.Font.Shading.BackgroundPatternColor = wdColorLightYellow
.Replacement.Text = ""
.Forward = True
.Wrap = wdFindContinue
.Execute Replace:=wdReplaceAll

End With
Selection.Find.Execute

End Sub
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