Help, Please, Table Borders LineWidth act as a Variant. Why Variant disregards my variant value?

Cendrinne · #1 09-05-2022, 12:22 AM

Hello pros,

I've succeeded to create a macro for Table Borders LineWidth to act as a Variant for the Last Row, for the bottom border.

Re: LineWidth = (sVar1), it remains at 1,50pt. Why?

Code:

   
Sub Table_1_Borders_LastRow_Top_050pt_Bottom_Var()
   Dim aTbl As Table
   Dim sVar1 As Variant               'Border size for the bottom row, bottom border.
   Dim LineWidth As Borders

Set aTbl = ActiveDocument.Tables(1)

   With Selection.Tables(1)
       .Rows.Last.Range.Select
   End With

    sVar1 = InputBox("Enter the Border size for the bottom row, on bottom border." _
      & vbCr & "Exemple en pt : 0.25, 0.50, 0.75, 1.00, 1.50, 2.25, 3.00, 4.50, 6.00.", "SUGGESTION", "3")

sVar1 = aTbl.Borders(wdBorderBottom).LineWidth

     With Selection.Range
        With .Borders(wdBorderTop)
            .LineStyle = wdLineStyleSingle
            .LineWidth = wdLineWidth050pt
            .Color = wdColorAutomatic
        End With

        With .Borders(wdBorderBottom)
            .LineStyle = wdLineStyleSingle
           '.LineWidth = wdLineWidth150pt
            .LineWidth = sVar1
            .Color = wdColorAutomatic
        End With
     End With
  
  On Error GoTo 0

End Sub

I've just edited my question, cause Trial and Error, I got it to work without error message. BUT, now the variant value, is Disregarded, and it puts 1.50 pt regardless the value I put as a variant.

What am I doing wrong, in your opinion?

Could any expert guide me please? I would be sooooo grateful.

Cendrinne

Italophile · #2 09-05-2022, 04:05 AM

You need to delete the line

Code:

 sVar1 = aTbl.Borders(wdBorderBottom).LineWidth

Cendrinne · #3 09-05-2022, 07:34 PM

Hello,
I've tried it, but the error message is pointing towards the line ==> .LineWidth = sVar1.

So is it a bug with Word, or my script?

But thanks for trying.

I'll continue trial and error, if I come up with an answer, I'll update the post.

Cendrinne

Cendrinne · #4 09-05-2022, 09:39 PM

Hello, pros,

I've tried anything I could think off. Now today, the width is 0.75" by default, regardless what I put.

I've even put xx, and no error message and got still 0.75".

I'm thinking, it cannot be done. Looking into wdLineWidth = Returns a WdLineWidth constant.
So I guess, a variable is not possible.

What do you all think?

Cendrinne

Guessed · #5 09-05-2022, 09:49 PM

wdLineWidth050pt has a value of 4
wdLineWidth150pt has a value of 12

So it appears you can't type in a value in points and expect Word to be able to assign that correctly to the LineWidth property. The only acceptable choices in that series of constants are:
2, 4, 6, 8, 12, 18, 24, 36, 48

You can review these constant values by using the VBA editor and going to View>Object Browser and searching for wdLineWidth which is the common prefix for all the possible relevant constants.

Cendrinne · #6 09-05-2022, 10:18 PM

Quote:

Originally Posted by Guessed

wdLineWidth050pt has a value of 4
wdLineWidth150pt has a value of 12

So it appears you can't type in a value in points and expect Word to be able to assign that correctly to the LineWidth property. The only acceptable choices in that series of constants are:
2, 4, 6, 8, 12, 18, 24, 36, 48

You can review these constant values by using the VBA editor and going to View>Object Browser and searching for wdLineWidth which is the common prefix for all the possible relevant constants.

So it is doable eventually, right?
But I did also added a Dim pt As Points and when I've entered a value as a Point system, I've put 1,50pt, but still got 0.75".

But thanks

Cendrinne

Guessed · #7 09-05-2022, 11:14 PM

Try this variation. It shows the ONLY possible values that can be entered and defaults to the same value as the first table in the document. I've coded it to apply all the same bottom row border settings as the first table in the document and allows you to override the bottom border setting that the first table had (if you wish to choose something else).
I did this because your code referenced the first table in the document and the first table in the selection. Those tables may or may not be the same thing.

Code:

Sub Table_Borders_LastRow()
  Dim aTbl As Table, aTblSel As Table
  Dim iLineWidth As Integer               'Border size for the bottom row, bottom border.
  
  Set aTbl = ActiveDocument.Tables(1)
  Set aTblSel = Selection.Tables(1)
  iLineWidth = aTbl.Borders(wdBorderBottom).LineWidth    'size of bottom border on first table
  
  iLineWidth = InputBox("Enter the Border size for the bottom row/bottom border." & vbCr & _
      "Choices are: 2, 4, 6, 8, 12, 18, 24, 36, 48", "SUGGESTION", iLineWidth)     'suggestion = first table's border size
  
  With aTblSel.Rows.Last
    .Borders = aTbl.Rows.Last.Borders
    .Borders(wdBorderBottom).LineWidth = iLineWidth
  End With
End Sub

Cendrinne · #8 09-05-2022, 11:45 PM

Quote:

Originally Posted by Guessed

Try this variation. It shows the ONLY possible values that can be entered and defaults to the same value as the first table in the document. I've coded it to apply all the same bottom row border settings as the first table in the document and allows you to override the bottom border setting that the first table had (if you wish to choose something else).
I did this because your code referenced the first table in the document and the first table in the selection. Those tables may or may not be the same thing.

Code:

Sub Table_Borders_LastRow()
  Dim aTbl As Table, aTblSel As Table
  Dim iLineWidth As Integer               'Border size for the bottom row, bottom border.
  
  Set aTbl = ActiveDocument.Tables(1)
  Set aTblSel = Selection.Tables(1)
  iLineWidth = aTbl.Borders(wdBorderBottom).LineWidth    'size of bottom border on first table
  
  iLineWidth = InputBox("Enter the Border size for the bottom row/bottom border." & vbCr & _
      "Choices are: 2, 4, 6, 8, 12, 18, 24, 36, 48", "SUGGESTION", iLineWidth)     'suggestion = first table's border size
  
  With aTblSel.Rows.Last
    .Borders = aTbl.Rows.Last.Borders
    .Borders(wdBorderBottom).LineWidth = iLineWidth
  End With
End Sub

Result:
I've tried it, however, it gave me an error message, Re the line:
.Borders(wdBorderBottom).LineWidth = iLineWidth.

I don't know how to fix it, however, I'm glad to have fixed my above script. I've even added on the same row a Top and Bottom Value as InputBoxes.

I've replaced ''As Variant'' for ''WdLineWidth'', Works great.

My script is maybe not as a pro like yourself would do it, but for No formal training, just analyzing scripts, I can say I'm getting not bad, right? Please give me a bone ''encouragement''

.

Here is my script:

Code:

   Dim aTbl As Table
   Dim sVar1 As WdLineWidth           'First Row - Top border
   Dim sVar2 As WdLineWidth           'First Row - Bottom border
   Dim LineWidth As Borders

Set aTbl = ActiveDocument.Tables(1)

    sVar1 = InputBox("For First Row, Enter the value on Top from the below numbers before the = symbole." _
      & vbCr & "Exemple : 2 =0.25/ 4 =0.50/ 6 =0.75/ 8 =1.00/" _
      & vbCr & "12 =1.50/ 18 =2.25/ 24 =3.00/ 36 =4.50/ 48 =6.00.", "SUGGESTION", "18")

    sVar2 = InputBox("For First Row, Enter the value on Bottom from the below numbers before the = symbole." _
      & vbCr & "Exemple : 2 =0.25/ 4 =0.50/ 6 =0.75/ 8 =1.00/" _
      & vbCr & "12 =1.50/ 18 =2.25/ 24 =3.00/ 36 =4.50/ 48 =6.00.", "SUGGESTION", "8")

   With Selection.Tables(1)
       .Rows.First.Range.Select
   End With

     With Selection.Range
       With .Borders(wdBorderTop)
            .LineStyle = wdLineStyleSingle
            .LineWidth = sVar1
            .Color = wdColorAutomatic
       End With

       With .Borders(wdBorderBottom)
            .LineStyle = wdLineStyleSingle
            .LineWidth = sVar2
            .Color = wdColorAutomatic
       End With
     End With
  
  On Error GoTo 0

And as the other person mentioned, sorry don't remember the person's name, I didn't need to put

Code:

 'sVar1 = aTbl.Borders(wdBorderTop).LineWidth

Note: I was reworking on the First Row, but I could easily modify it for the Bottom Row, like your script and my original post.

Once I know how to do this, I could now try to do it to all tables in a document. At times, I have over 80 tables in a document, which takes me much times, to fix them to be constant.

From the bottom of my heart, thank you for guiding me and helping me or anyone with those type of script's

Cendrinne

Cendrinne · #9 09-06-2022, 12:10 AM

Quote:

Originally Posted by Italophile

You need to delete the line

Code:

 sVar1 = aTbl.Borders(wdBorderBottom).LineWidth

It was more than that suggestion, but I did keep your suggestion in mind, and it contributed to my script.

Thanks for your input by the way

Cendrinne

Cendrinne · #10 09-06-2022, 12:23 AM

Here is my script for all tables in a document. Had to rearrange the order of certain things. But this below works for me

For this bellow script, I've created for the Last Row:

Code:

 
   Dim xTbl As Table
   Dim sVar1 As WdLineWidth           'For the Last Row: Value to put as Top Border
   Dim sVar2 As WdLineWidth           'For the Last Row: Value to put as Bottom Border
   Dim LineWidth As Borders
    
    sVar1 = InputBox("Last Row - Enter the size value from the below choices before the symbole = for the Top Border." _
      & vbCr & "Exemple : 2 =0.25/ 4 =0.50/ 6 =0.75/ 8 =1.00/" _
      & vbCr & "12 =1.50/ 18 =2.25/ 24 =3.00/ 36 =4.50/ 48 =6.00.", "SUGGESTION", "8")

    sVar2 = InputBox("Last Row - Enter the size value from the below choices before the symbole = for the Bottom Border." _
      & vbCr & "Exemple : 2 =0.25/ 4 =0.50/ 6 =0.75/ 8 =1.00/" _
      & vbCr & "12 =1.50/ 18 =2.25/ 24 =3.00/ 36 =4.50/ 48 =6.00.", "SUGGESTION", "18")
    
    For Each xTbl In ActiveDocument.Tables
      xTbl.Rows.Last.Range.Select

     With Selection.Range
       With .Borders(wdBorderTop)
            .LineStyle = wdLineStyleSingle
            .LineWidth = sVar1
            .Color = wdColorAutomatic
       End With

       With .Borders(wdBorderBottom)
            .LineStyle = wdLineStyleSingle
            .LineWidth = sVar2
            .Color = wdColorAutomatic
       End With
     End With
  
   Next xTbl
  
  On Error GoTo 0

Hopefully it works for others on their PC's.

Cendrinne

gmaxey · #11 09-11-2022, 04:13 AM

You could employ a simple userform with two listboxes and a command button. Let prone to user errors:

Code:

Option Explicit
Private Sub UserForm_Initialize()
 With ListBox1
   .AddItem
   .List(.ListCount - 1, 0) = "0.25 pt"
   .List(.ListCount - 1, 1) = 2
   .AddItem
   .List(.ListCount - 1, 0) = "0.50 pt"
   .List(.ListCount - 1, 1) = 4
   .AddItem
   .List(.ListCount - 1, 0) = "0.75 pt"
   .List(.ListCount - 1, 1) = 6
   .AddItem
   .List(.ListCount - 1, 0) = "1 pt"
   .List(.ListCount - 1, 1) = 8
   .AddItem
   .List(.ListCount - 1, 0) = "1.5 pt"
   .List(.ListCount - 1, 1) = 12
   .AddItem
   .List(.ListCount - 1, 0) = "2.25 pt"
   .List(.ListCount - 1, 1) = 18
   .AddItem
   .List(.ListCount - 1, 0) = "3.00 pt"
   .List(.ListCount - 1, 1) = 24
   .AddItem
   .List(.ListCount - 1, 0) = "4.5 pt"
   .List(.ListCount - 1, 1) = 36
   .AddItem
   .List(.ListCount - 1, 0) = "6 pt"
   .List(.ListCount - 1, 1) = 48
   .ListIndex = 3
  End With
  With ListBox2
    .List = ListBox1.List
    .ListIndex = 5
  End With
lbl_Exit:
  Exit Sub
End Sub

Private Sub CommandButton1_Click()
Dim oTbl As Table
  For Each oTbl In ActiveDocument.Tables
    oTbl.Rows.Last.Range.Select
    With Selection.Range
      With .Borders(wdBorderTop)
        .LineStyle = wdLineStyleSingle
        .LineWidth = ListBox1.Column(1)
        .Color = wdColorAutomatic
      End With
      With .Borders(wdBorderBottom)
        .LineStyle = wdLineStyleSingle
        .LineWidth = ListBox2.Column(1)
        .Color = wdColorAutomatic
      End With
    End With
  Next oTbl
lbl_Exit:
  Exit Sub
End Sub

Cendrinne · #12 09-12-2022, 10:51 PM

Quote:

Originally Posted by gmaxey

You could employ a simple userform with two listboxes and a command button. Let prone to user errors:

Code:

Option Explicit
Private Sub UserForm_Initialize()
 With ListBox1
   .AddItem
   .List(.ListCount - 1, 0) = "0.25 pt"
   .List(.ListCount - 1, 1) = 2
   .AddItem
   .List(.ListCount - 1, 0) = "0.50 pt"
   .List(.ListCount - 1, 1) = 4
   .AddItem
   .List(.ListCount - 1, 0) = "0.75 pt"
   .List(.ListCount - 1, 1) = 6
   .AddItem
   .List(.ListCount - 1, 0) = "1 pt"
   .List(.ListCount - 1, 1) = 8
   .AddItem
   .List(.ListCount - 1, 0) = "1.5 pt"
   .List(.ListCount - 1, 1) = 12
   .AddItem
   .List(.ListCount - 1, 0) = "2.25 pt"
   .List(.ListCount - 1, 1) = 18
   .AddItem
   .List(.ListCount - 1, 0) = "3.00 pt"
   .List(.ListCount - 1, 1) = 24
   .AddItem
   .List(.ListCount - 1, 0) = "4.5 pt"
   .List(.ListCount - 1, 1) = 36
   .AddItem
   .List(.ListCount - 1, 0) = "6 pt"
   .List(.ListCount - 1, 1) = 48
   .ListIndex = 3
  End With
  With ListBox2
    .List = ListBox1.List
    .ListIndex = 5
  End With
lbl_Exit:
  Exit Sub
End Sub

Private Sub CommandButton1_Click()
Dim oTbl As Table
  For Each oTbl In ActiveDocument.Tables
    oTbl.Rows.Last.Range.Select
    With Selection.Range
      With .Borders(wdBorderTop)
        .LineStyle = wdLineStyleSingle
        .LineWidth = ListBox1.Column(1)
        .Color = wdColorAutomatic
      End With
      With .Borders(wdBorderBottom)
        .LineStyle = wdLineStyleSingle
        .LineWidth = ListBox2.Column(1)
        .Color = wdColorAutomatic
      End With
    End With
  Next oTbl
lbl_Exit:
  Exit Sub
End Sub

OK, I'll look at it tomorrow.
Just came to check something, but must go to bed for work.

Night night. Nice hearing from you

Cendrinne · #13 09-13-2022, 11:36 PM

Quote:

Originally Posted by gmaxey

You could employ a simple userform with two listboxes and a command button. Let prone to user errors:

Code:

Option Explicit
Private Sub UserForm_Initialize()
 With ListBox1
   .AddItem
   .List(.ListCount - 1, 0) = "0.25 pt"
   .List(.ListCount - 1, 1) = 2
   .AddItem
   .List(.ListCount - 1, 0) = "0.50 pt"
   .List(.ListCount - 1, 1) = 4
   .AddItem
   .List(.ListCount - 1, 0) = "0.75 pt"
   .List(.ListCount - 1, 1) = 6
   .AddItem
   .List(.ListCount - 1, 0) = "1 pt"
   .List(.ListCount - 1, 1) = 8
   .AddItem
   .List(.ListCount - 1, 0) = "1.5 pt"
   .List(.ListCount - 1, 1) = 12
   .AddItem
   .List(.ListCount - 1, 0) = "2.25 pt"
   .List(.ListCount - 1, 1) = 18
   .AddItem
   .List(.ListCount - 1, 0) = "3.00 pt"
   .List(.ListCount - 1, 1) = 24
   .AddItem
   .List(.ListCount - 1, 0) = "4.5 pt"
   .List(.ListCount - 1, 1) = 36
   .AddItem
   .List(.ListCount - 1, 0) = "6 pt"
   .List(.ListCount - 1, 1) = 48
   .ListIndex = 3
  End With
  With ListBox2
    .List = ListBox1.List
    .ListIndex = 5
  End With
lbl_Exit:
  Exit Sub
End Sub

Private Sub CommandButton1_Click()
Dim oTbl As Table
  For Each oTbl In ActiveDocument.Tables
    oTbl.Rows.Last.Range.Select
    With Selection.Range
      With .Borders(wdBorderTop)
        .LineStyle = wdLineStyleSingle
        .LineWidth = ListBox1.Column(1)
        .Color = wdColorAutomatic
      End With
      With .Borders(wdBorderBottom)
        .LineStyle = wdLineStyleSingle
        .LineWidth = ListBox2.Column(1)
        .Color = wdColorAutomatic
      End With
    End With
  Next oTbl
lbl_Exit:
  Exit Sub
End Sub

You are WAY more advance than my level of knowledge. Wow, Wish you would make a book for less familiar people. Even stuff on your website, I'm not certain to understand lot's of contents.

Re this script,
I get an error message, and it highlights ListBox. It say's it's undefined Variable.
So I've tried to put Dim ListBox1 As ListBox, ListBox2 As ListBox = Worked but now it flags .AddItem

So I wish to have tried it, but I need more guidance to figure out what it's supposed to do.

But I'm truly grateful for your wanting to help me

Got to go to bed for now.

Night night

gmaxey · #14 09-14-2022, 01:28 AM

You have to create a userform with two listboxes and a command button. The code goes in the userform module. See Create & Employ a Userform

Cendrinne · #15 09-14-2022, 07:15 PM

Quote:

Originally Posted by gmaxey

You have to create a userform with two listboxes and a command button. The code goes in the userform module. See Create & Employ a Userform

I can't wait to try that this week-end. I like to learn new ways.
I'll let you know for sure.

Thanks Greg.

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