[Doug Robbins - Word MVP]

The following code works fine for me:

Dim i As Long
Dim myTable As Table
Dim myrange As Range
With ActiveDocument
For i = 1 To .Tables.Count
Set myTable = .Tables(i)
With myTable
Set myrange = .Range
Select Case myrange.Information(wdActiveEndPageNumber)
Case Is 99 201
.AutoFitBehavior wdAutoFitFixed
.Columns(1).Width = CentimetersToPoints(1.78)
.Columns(2).Width = CentimetersToPoints(6.37)
.Columns(3).Width = CentimetersToPoints(6.37)
.Columns(4).Width = CentimetersToPoints(1.78)
End Select
End With
Next i
End With


--
Hope this helps.

Please reply to the newsgroup unless you wish to avail yourself of my
services on a paid consulting basis.

Doug Robbins - Word MVP, originally posted via msnews.microsoft.com
"Rage" wrote in message
...
It's no where near 60cm - that's huge - the first and last columns are
1.78
cm long, and the 2nd and 3rd columns are 6.37 cm long, giving a total of
16.3
cm for the table width (which is the same width as my A4 word document not
including margins)...

"Doug Robbins - Word MVP" wrote:

So you want to make the total width of the table 60cm. That would exceed
the maximum page width that Word can accommodate which is abot 55.88cm.

There is nothing wrong with the macro as long an you do not have it
contain
values that would make the table wider than the maximum page width.

--
Hope this helps.

Please reply to the newsgroup unless you wish to avail yourself of my
services on a paid consulting basis.

Doug Robbins - Word MVP, originally posted via msnews.microsoft.com
"Rage" wrote in message
...
4: the four columns are split up into groups of 2, so columns 1 & 4 are
the
same size, and columns 2 & 3 are the same size too.

Cheers

"Doug Robbins - Word MVP" wrote:

How many columns are there in your table?

--
Hope this helps.

Please reply to the newsgroup unless you wish to avail yourself of my
services on a paid consulting basis.

Doug Robbins - Word MVP, originally posted via msnews.microsoft.com
"Rage" wrote in message
...
Hi Doug,

Thanks for your reply! There's only one thing wrong:

I get a:

Run-time error '5941':

The requested member of the collection does not exist.

at the line:
.Columns(3).Width = CentimetersToPoints(6.37)

(I did take out the apostrophe).

Any ideas?

Cheers.





"Doug Robbins - Word MVP" wrote:

Use:

Dim i As Long
Dim myTable As Table
Dim myrange As Range
With ActiveDocument
For i = 1 To .Tables.Count
Set myTable = .Tables(i)
With myTable
Set myrange = .Range
Select Case myrange.Information(wdActiveEndPageNumber)
Case Is 99 201
.AutoFitBehavior wdAutoFitFixed
.Columns(1).Width = CentimetersToPoints(10)
.Columns(2).Width = CentimetersToPoints(2)
'.Columns(3).Width = CentimetersToPoints(#)
'.Columns(4).Width = CentimetersToPoints(#)
End Select
End With
Next i
End With


--
Hope this helps.

Please reply to the newsgroup unless you wish to avail yourself of
my
services on a paid consulting basis.

Doug Robbins - Word MVP, originally posted via msnews.microsoft.com
"Rage" wrote in message
...
Hi gang,

I have a 300-odd MS Word Document (Word Version 2003) with quite
a
lot
of
tables, and I was wondering - can someone possibly provide me
with a
macro
which will take all the tables in a range (e.g.) pages 100 - 200,
and
set
the
width of each column to a specific value (e.g.) column 1 = 10 cm,
column 2
=
20 cm etc.?

All the tables in the range needed have 4 columns, but each
column
needs
to
be a different value to the others.

Thanks in advance,

Rage.