Word Macro to find LeftIndent

[skbaskar]

Hi All,
I am new to vb and i tried a macro in word to find the paragraph indents and insert some tag for that indented text for e.g. indent 0.5 to 0.75 to be coded as <indent1> and 0.76 to 1.00 as “<indent2>”

Accommodate, (66) p.22, p.24,p.55
Acid
<indent1>Diphos phonic Acid, (35)p.36
<indent2>Acryli c Acid, (35) p.36-37

And this is my code

[CODE=vb]Sub IndentTagging()
Indent = Val(0.5)
With Selection.Find. ParagraphFormat
Do While Indent <= InchesToPoints( 0.75)
.LeftIndent = InchesToPoints( Indent)
MsgBox ("this is left indent " & (.LeftIndent))
With Selection.Find
.Text = ""
.Replacement.Te xt = "<indent1>^ &"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchWildcards = False
End With
Selection.Find. Execute Replace:=wdRepl aceAll
Indent = Indent + 0.01
MsgBox ("this is variable " & InchesToPoints( Indent))
Loop
End With
End Sub[/CODE]

The problem is when I pass the variable value to Leftindent it rounded the value i.e., 44.44 converted as 44.45 please tell me how to avoid this

Thanks in Advance
Baskar

[Killer42]

This may simply be a limitation inherent in the system. there are 72 points per inch, so presumably the InchesToPoints function cannot return every value between .0 and .99, but probably jumps to the nearest 100th.

Or I could be completely wrong, of course. Try passing all the values from 0.00 to 0.99 to the function, and see what you get back.

[skbaskar]

This may simply be a limitation inherent in the system. there are 72 points per inch, so presumably the InchesToPoints function cannot return every value between .0 and .99, but probably jumps to the nearest 100th.

Or I could be completely wrong, of course. Try passing all the values from 0.00 to 0.99 to the function, and see what you get back.

Hi Killer,

First thanks for your reply and you said InchesToPoints function cannot return every value between .0 and .99 but in my code the second message box returning the values .0 to .99 even i used the InchesToPoints function and when i pass the value to .LeftIndent then only it round the number. Till i dont know how to controle it.

Sub IndentTagging()
Indent = Val(0.5)
With Selection.Find. ParagraphFormat
Do While Indent <= InchesToPoints( 0.75)
.LeftIndent = InchesToPoints( Indent)
MsgBox ("this is left indent " & (.LeftIndent))
With Selection.Find
.Text = ""
.Replacement.Te xt = "<indent1>^ &"
.Forward = True
.Wrap = wdFindContinue
.Format = True
.MatchWildcards = False
End With
Selection.Find. Execute Replace:=wdRepl aceAll
Indent = Indent + 0.01
MsgBox ("this is variable " & InchesToPoints(Indent)) 'This message box showing the 0.0 to 0.99 values
Loop
End With
End Sub

Thanks
Baskar K.

[Killer42]

Have you checked the documentation for LeftIndent? It's conceivable that it only accepts whole point values, regardless of what the conversion function can produce. Or something.

It looks as though LeftIndent doesn't work in my old version of Excel here at home, so I'll have to have a look in the morning when I get back to work.