Any helper re setting tab stops (thank you!)

[singeredel]

Please see my last two posts under "Set tabs based on distance from right
margin." I am trying to set two tabs a specific distance from the right
margin based on the length of string variables, but I get a run-time error on
the code at the end when setting the tabs that the measurement must be
between -1584 pt and 1584 pt. Do I need to convert my tab calculations
somehow?

Thank you!

 

[Helmut Weber]

Hi Julie,

you'd better stay in the same thread,
unless it isn't weeks old.

First, a string variable hasn't a physical length at all.
It is something abstract.

Second, if you think all characters in a monospaced font
would have always the same length on the screen, you are wrong.
Create a new paragraph, type "XXXXXXXXXXXXXXXXXXX",
put the cursor at the paragraph's start, and try the following:

Dim l As Long
Dim p1 As Single
Dim p2 As Single
For l = 1 To 12
p1 = Selection.Information(wdHorizontalPositionRelativeToPage)
Selection.MoveRight
p2 = Selection.Information(wdHorizontalPositionRelativeToPage)
MsgBox p2 - p1
Next

I get results from 7.34 to 6.75 !

From that I deduce that only an approximate solution
will be possible. Maybe quite other ways could be recommended,
like putting the text in a table and format the rightmost cell
as leftaligned.

For calculating the position of the tabstop,
whereby to find the right approximation factor is up to you:

Dim sTm As String ' temporary string
Dim sRM As Single ' right margin
Dim sPW As Single ' page width
Dim sGt As Single ' gutter width
Dim sPs As Single ' position of tabstop
With ActiveDocument.PageSetup
sRM = .RightMargin
sPW = .PageWidth
sGt = .Gutter
End With
sPs = sPW - sGt - sRM
sTm = "Have some fun"
' deduce from tab position on left margin
' the approximately lenght of the string
sPs = sPs - Len(sTm) * 7.5 * ApproxFactor ' ???
Selection.Paragraphs(1).TabStops.Add Position:=sPs
' Plus some unknown constants, possibly

Greetings from Bavaria, Germany

Helmut Weber, MVP
"red.sys" & chr(64) & "t-online.de"
Word XP, Win 98
http://word.mvps.org/

 

[singeredel]

Hi Helmut,

Thank you for your reply. I do not understand this "approximate factor."
When I ran your code, I got 7.13 to 7.19, but I don't know what this means.
Also, I ran your code for setting the tab, and it set it off the page. What
is the "7.5" number you have in your sample code? This should be used plus
the approximate factor? I tried using a number 7 where you had
"ApproxFactor", but I ended up with no tab at all.

Thanks from Julie

 

[Helmut Weber]

Hi Julie,

When I ran your code, I got 7.13 to 7.19, but I don't know what this means.

7.13 to 7.19 would be the width of a character in points,
according to the measuring method applied,
which seems to be constantly 6, by the way,
for courier new 10 point size with zoom factor 200 percent.

I do not understand this "approximate factor."

If you get different values for the width of a character,
you have to decide, whether you want to use the minimum width,
maximum width or the average or just add or deduct some constant
value from the possibly varying width of the complete string.
You might also call it a "corrective".

What was missing in my calculation was the left margin. :-(

So we got this, not regarding gutter width, which I left out:

Sub CalcTabPositionCell()
Dim sTm As String ' temporary string
Dim sRM As Single ' right margin
Dim sLM As Single ' left margin
Dim sPW As Single ' page width
Dim sPs As Single ' position of tabstop
With ActiveDocument.PageSetup
sLM = .LeftMargin
sRM = .RightMargin ' 70
sPW = .PageWidth ' 595
End With
sPs = sPW - (sLM + sRM)
sTm = "To be or not to be"
' deduce from tab position on left margin
' the approximately lenght of the string
sPs = sPs - (Len(sTm)) * 6 - 2 ' corrective 2
Selection.Paragraphs(1).TabStops.Add _
Position:=sPs, _
Alignment:=wdAlignTabLeft
End Sub

which puts a tabstop left from the right margin,
just left enough to hold "To be or not to be".
Given that a character is 6 points wide plus 2 points
to correct fractions or inconsistencies in the calculation.

Tested with zoom 200 percent
on 21 cm wide paper (DIN A4),
left margin = 2.5 cm
right margin = 2.5 cm
no gutter

Greetings from Bavaria, Germany

Helmut Weber, MVP
"red.sys" & chr(64) & "t-online.de"
Word XP, Win 98
http://word.mvps.org/

 

[singeredel]

Thank you so much. Your test was successful using your string sTm; however,
when I applied your code using my variables, for some reason it sets the tab
at 6.31", which does not accommodate the length of the variable with a margin
of 6.5". I am sorry to try your patience. Perhaps you would be kind enough to
look at my latest version and see if you spot something I am doing wrong. My
additions are indented. I never though this would be that difficult! :O

PatientName$ = "Smith, Joe"
SSN$ = "555-55-5555"
AttentionLine$ = "John Doe"
DOE$ = "02/28/05"


Dim sTm As String ' temporary string
Dim sRM As Single ' right margin
Dim sLM As Single ' left margin
Dim sPW As Single ' page width
Dim sPs As Single ' position of tabstop
Dim sLen As Single
Dim sLen1 As Single
Dim sLen2 As Single
Dim sLen3 As Single
Dim sLen4 As Single

sLen1 = Len(PatientName$)
sLen2 = Len(SSN$)
sLen3 = Len(AttentionLine$)
sLen4 = Len(DOE$)


If sLen1 >= sLen2 And sLen1 >= sLen3 And sLen1 >= sLen4 Then sTm =
sLen1
If sLen2 >= sLen1 And sLen2 >= sLen3 And sLen2 >= sLen4 Then sTm =
sLen2
If sLen3 >= sLen1 And sLen3 >= sLen2 And sLen3 >= sLen4 Then sTm =
sLen3
If sLen4 >= sLen1 And sLen4 >= sLen2 And sLen4 >= sLen3 Then sTm =
sLen4

With ActiveDocument.PageSetup
sLM = .LeftMargin
sRM = .RightMargin ' 72
sPW = .PageWidth ' 612
End With

sPs = sPW - (sLM + sRM)
'sTm = "To be or not to be"
' deduce from tab position on left margin
' the approximately lenght of the string
sPs = sPs - (Len(sTm)) * 6 - 2 ' corrective 2
Selection.Paragraphs(1).TabStops.Add _
Position:=sPs, _
Alignment:=wdAlignTabLeft

Julie

 

[Helmut Weber]

Hi Julie,

with a sense of humor,

I wish I never had answered before (not really) ;-)
Now my obligation seems to be to finish it.
I don't know whether I can do it.

To the co-readers: Any other input is most welcome!

Are you using a monospaced font?!
What is all that

If sLen1 >= sLen2 And sLen1 >= sLen3 And sLen1 >= sLen4 Then

about?

Oh, I'm seeing something.
You are assignign a variable of type single to a string. !!!

Dim sTm As String ' temporary string
Dim sLen4 As Single

If sLen4 >= sLen1 [...] sLen4 >= sLen3 Then sTm = sLen4

Prone to cause problems.

Greetings from Bavaria, Germany

Helmut Weber, MVP
"red.sys" & chr(64) & "t-online.de"
Word XP, Win 98
http://word.mvps.org/

 

[singeredel]

Okay, this is what I am looking for as a final result:

DEPARTMENT OF SOCIAL SERVICES RE: SMITH, JOE
Disability and Adult Programs SSN: 555-55-5555
P.O. Box 60999 ATTN: JOHN DOE
Los Angeles, CA 90060

The length of the data of the three items above called "RE:", "SSN:" and
"ATTN:" (designated as PatientName$, SSN$, and AttentionLine$) needs to be
determined so that all three of those lines can have two tabs set, one
left-aligned tab to accommodate the longest of the three items which needs to
line up with the right margin, and a second right-justified tab to line up
RE:, SSN:, AND ATTN: This fluctuates depending on which of the three items is
the longest. There are bookmarks at these three locations to insert the
variable data.

I see the variable sTm was assigned as a String instead of a Single.
However, this did not seem to make any difference after it was changed. The
first tab stop showed up at 6.14", which still does not accommodate the
longest variable data.

I hope this helps to clear up what I am trying to do. I know I have been a
pain, but your help is very much appreciated. Thank you!

Hi Julie,

with a sense of humor,

I wish I never had answered before (not really) ;-)
Now my obligation seems to be to finish it.
I don't know whether I can do it.

To the co-readers: Any other input is most welcome!

Are you using a monospaced font?!
What is all that

If sLen1 >= sLen2 And sLen1 >= sLen3 And sLen1 >= sLen4 Then

about?

Oh, I'm seeing something.
You are assignign a variable of type single to a string. !!!

Dim sTm As String ' temporary string
Dim sLen4 As Single

If sLen4 >= sLen1 [...] sLen4 >= sLen3 Then sTm = sLen4

Prone to cause problems.

Greetings from Bavaria, Germany

Helmut Weber, MVP
"red.sys" & chr(64) & "t-online.de"
Word XP, Win 98

 

[singeredel]

Helmut -- Finally success! I discovered what the problem was. In your code
you had:
sPs = sPs - (Len(sTm)) * 6 - 2 ' corrective 2
sTm is already a length, and this was taking the length of a length. So I
removed the Len(sTm) and made it just sTm, or:
sPs = sPs - sTm * 6 - 2 ' corrective 2
But you got me there! Thank you!

Julie

Hi Julie,

with a sense of humor,

I wish I never had answered before (not really) ;-)
Now my obligation seems to be to finish it.
I don't know whether I can do it.

To the co-readers: Any other input is most welcome!

Are you using a monospaced font?!
What is all that

If sLen1 >= sLen2 And sLen1 >= sLen3 And sLen1 >= sLen4 Then

about?

Oh, I'm seeing something.
You are assignign a variable of type single to a string. !!!

Dim sTm As String ' temporary string
Dim sLen4 As Single

If sLen4 >= sLen1 [...] sLen4 >= sLen3 Then sTm = sLen4

Prone to cause problems.

Greetings from Bavaria, Germany

Helmut Weber, MVP
"red.sys" & chr(64) & "t-online.de"
Word XP, Win 98