Table formatting- column widths to percent

dherr · #1 03-17-2015, 10:03 AM

Hi,
I want to set the columns of a table to procentual values and have written the following code:

Code:

Sub setProz(tb As Word.table, prozentual As Variant)
    Dim i As Integer, co As Integer
    With tb
        .PreferredWidthType = wdPreferredWidthPercent
        co = .columns.Count
        For i = 0 To co - 1
            .columns(i + 1).PreferredWidth = Val(prozentual(i))
        Next
    End With
End Sub

The parameters are first a table, second an array, for instance
Array(10, 50, 20, 20) when the table has 4 columns.
The problem is when I use this code, the table doesn't change the column widths at normal runtime.
When I set a debugging breakpoint to the line 'End Sub' the table changes after a second.
I tried to 'refresh' or 'update' the table, but no success...
What's to do?

Regards-
Dietrich

gmaxey · #2 03-17-2015, 03:22 PM

I called your code using the following and it worked fine:

Code:

Sub Test()
  Dim arrPercent() As String
  arrPercent = Split("10,50,20,20", ",")
  setProz ActiveDocument.Tables(1), arrPercent
End Sub

dherr · #3 03-17-2015, 03:50 PM

Thanks for your answer, Greg,
when I use your advice I have the same effect. So my code is OK.

Again: the problem is, that the table shows right only then when I set this breakpoint during debugging; I think because of the pause (the table has 300 rows). When I delete this breakpoint the table changes NOT to the new column widths.
It seems to me that because of the breakpoint the table 'has time' to 'refresh' and shows new columns width. Otherwise, without breakpoint, not.

Something new:
Today I have set a STOP (VBA order, not a breakpoint) after my setProz routine and the table changes the column widths... astonishing.
But now I have a new problem: How to continue the code after the STOP ???

Regards
Dietrich

adamyawn · #4 05-02-2023, 09:37 AM

I know this post was created ages ago, but it's worth noting that this line

.Columns(i + 1).PreferredWidth = ...

works better as

.Columns(i + 1).Width = ...

in terms of sizing each column. Thank you for sharing this code with everyone!

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