Conditional tied to trying to find a " " space character

Dart82 · #1 10-03-2013, 10:16 AM

Hey, I'm trying to write MS word VBA code to find a space to the left of the cursor. If the space is found, go to right of the space, otherwise, type a space. I've also attached an image to try to be more specific.

My code assumes I start to the left of the required location. I then select the text one character to the left. I search this text for a " ". However, I can't get it to work correctly.

Could anybody please help me get my code, shown below, sorted?

Code:

Sub IfSpaceFound()
    Selection.MoveLeft Unit:=wdCharacter, Count:=1, Extend:=wdExtend
    With Selection.Find
        .ClearFormatting
        .Text = " "
        .Forward = True
        .Wrap = False
        .Format = True
        .MatchCase = False
        .MatchWholeWord = False
        .MatchWildcards = True
        .MatchSoundsLike = False
        .MatchAllWordForms = False
        .Execute
        If .Found
            Selection.MoveRight Unit:=wdCharacter, Count:=1
        Else
            Selection.TypeText Text:=" "
        End If
    End With
End Sub

macropod · #2 10-03-2013, 02:47 PM

Try:

Code:

Sub IfSpaceFound()
With Selection.Characters.First
  If .Start = ActiveDocument.Range.Start Then
    .InsertBefore " "
    .MoveStart wdCharacter, 1
  ElseIf .Previous <> " " Then
    .InsertBefore " "
    .MoveStart wdCharacter, 1
  End If
End With
End Sub

slaycock · #3 10-04-2013, 08:13 AM

Why are you trying to do this? Surely it would be easier to do a search and replace on the document e.g. .^p for . ^p.

If you want to do it programatically then your code will work if you change to count:= -1. So if you are in the no space case in your thumbnail, the count:= -1 will select to the left of the cursor. if you use count:=1 then you select to the right of the cursor.

macropod · #4 10-04-2013, 01:47 PM

Slaycock,

The limitation of your approach is that it will only work at paragraph ends and even then only of the paragraph break is preceded by a period. It won't work in any other situation. The code I posted will work anywhere in a document - including at the very start, where your '-1' suggestion would cause an error.

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