DOCUMENT:Q217012  11-JAN-2001  [vbwin]
TITLE   :HOWTO: Format Strings to Right-Justify When Printing
PRODUCT :Microsoft Visual Basic for Windows
PROD/VER:WINDOWS:5.0,6.0
OPER/SYS:
KEYWORDS:kbString kbVBp kbVBp500 kbVBp600 kbGrpDSVB kbDSupport

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The information in this article applies to:

 - Microsoft Visual Basic Learning Edition for Windows, versions 5.0, 6.0 
 - Microsoft Visual Basic Professional Edition for Windows, versions 5.0, 6.0 
 - Microsoft Visual Basic Enterprise Edition for Windows, versions 5.0, 6.0 
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SUMMARY
=======

There are several different ways to right-justify strings using the Format
function:

 - Use the @ character.

 - Use the RSet function.

 - Use workarounds with the Format$ function.

MORE INFORMATION
================

Using the @ character::

NOTE: This technique is only effective with monospace fonts, such as Courier
New.

1. Format the number into a string with numeric conversion characters, for
   example, $##0.00.

2. Format the resulting string with a format string consisting of a number of @
   characters equal in length to the desired format, for example, @@@@@@@.

The following code sample formats several numbers using seven @ characters and a
seven character format, $##0.00.

      Print "|" & Format$(Format$(1.5, "$##0.00"), "@@@@@@@") & "|"
      Print "|" & Format$(Format$(12.5, "$##0.00"), "@@@@@@@") & "|"
      Print "|" & Format$(Format$(123.5, "$##0.00"), "@@@@@@@") & "|"

The output is;

|  $1.50|
| $12.50|
|$123.50|

Using the RSet function::

When used in conjunction with RSet, the format function works on fixed length
strings. The following code sample illustrates the use of RSet:

      x = (Format$(123.5, "$##0.00"))
      Print "x" & x & "x"
      RSet x = (Format$(1.5, "$##0.00"))
      Print "x" & x & "x"

The output is:

x$123.50x 
x  $1.50x

Workarounds using the Format$ function::

NOTE: These techniques are only effective with monospace fonts, such as Courier
New.

The Format$ function does not right-justify strings when used with the # symbol.
The first code sample uses the Len function to determine how many spaces need to
be added to the left of the string representing the number, in order to right
justify the string:

      required = 8 ' longest number expected
      a = 1.23
      b = 44.56
      num1$ = Format$(a, "#0.00") ' this converts the number to a string
      num2$ = Format$(b, "#0.00") ' with 2 decimal places and a leading zero
      'Debug.Print num2$
      If (required - Len(num1$)) > 0 Then
      num1$ = Space$(required - Len(num1$)) + num1$
      End If
      
      If (required - Len(num2$)) > 0 Then
      num2$ = Space$(required - Len(num2$)) + num2$
      End If
   ' test output
      Print num1$
      Print num2$

The output is:

   1.23
  44.56

The second Format$ sample is reprinted with the permission of its author, Karl
Peterson. His LPad function uses the Right$ function:

   Private Function LPad(ValIn As Variant, nDec As Integer, _
                         WidthOut As Integer) As String
   '
   ' Formatting function left pads with spaces, using specified
   ' number of decimal digits.
   '
      If IsNumeric(ValIn) Then
         If nDec > 0 Then
            LPad = Right$(Space$(WidthOut) & _
                   Format$(ValIn, "0." & String$(nDec, "0")), _
                   WidthOut)
         Else
            LPad = Right$(Space$(WidthOut) & Format$(ValIn, "0"), WidthOut)
         End If
      Else
         LPad = Right$(Space$(WidthOut) & ValIn, WidthOut)
      End If
   End Function

Step by Step Sample:

1. Start a new Visual Basic Standard EXE project. Form1 is created by default.

2. Add four CommandButton controls to Form1. Position them to the far right of
   the form window.

3. Add the following code to the General Declarations section of Form1:

   Option Explicit

   Private Sub Command1_Click()
      Me.Print "|" & Format$(Format$(1.5, "$##0.00"), "@@@@@@@") & "|"
      Me.Print "|" & Format$(Format$(12.5, "$##0.00"), "@@@@@@@") & "|"
      Me.Print "|" & Format$(Format$(123.5, "$##0.00"), "@@@@@@@") & "|"
   End Sub

   Private Sub Command2_Click()
      Dim x As String
      x = (Format$(123.5, "$##0.00"))
      Me.Print "x" & x & "x"
      RSet x = (Format$(1.5, "$##0.00"))
      Me.Print "x" & x & "x"
   End Sub

   Private Sub Command3_Click()
      Dim required As Integer
      Dim a As Single
      Dim b As Single
      Dim num1$, num2$

      required = 8 ' longest number expected
      a = 1.23
      b = 44.56
      num1$ = Format$(a, "#0.00") ' this converts the number to a string
      num2$ = Format$(b, "#0.00") ' with two decimal places and a leading zero
      'Debug.Print num2$
      If (required - Len(num1$)) > 0 Then
         num1$ = Space$(required - Len(num1$)) & num1$
      End If
      
      If (required - Len(num2$)) > 0 Then
         num2$ = Space$(required - Len(num2$)) & num2$
      End If
   ' test output
      Me.Print num1$
      Me.Print num2$
   End Sub

   Private Sub Command4_Click()
      Dim xstring As String
      xstring = LPad(2.3, 2, 7)
      Me.Print "K" & xstring & "K"
   End Sub

   Private Sub Form_Load()
      Command1.Caption = "@"
      Command1.Font.Size = 18
      Command2.Caption = "Rset"
      Command3.Caption = "Format$"
      Command4.Caption = "VBPJ"
      Me.Font.Name = "Courier New"
   End Sub

   Private Function LPad(ValIn As Variant, nDec As Integer, _
                         WidthOut As Integer) As String
   '
   ' Formatting function left pads with spaces, using specified
   ' number of decimal digits.
   '
      If IsNumeric(ValIn) Then
         If nDec > 0 Then
            LPad = Right$(Space$(WidthOut) & _
                   Format$(ValIn, "0." & String$(nDec, "0")), _
                   WidthOut)
         Else
            LPad = Right$(Space$(WidthOut) & Format$(ValIn, "0"), WidthOut)
         End If
      Else
         LPad = Right$(Space$(WidthOut) & ValIn, WidthOut)
      End If
   End Function

4. Run the program, click the command buttons, and observe the results.

REFERENCES
==========

   Q95945 How to Right Justify Numbers Using Format$

   Q79094 PRB: Format$ Using # for Digit Affects Right Alignment

Additional query words: right-justify align alignment column

======================================================================
Keywords          : kbString kbVBp kbVBp500 kbVBp600 kbGrpDSVB kbDSupport 
Technology        : kbVBSearch kbAudDeveloper kbZNotKeyword6 kbZNotKeyword2 kbVB500Search kbVB600Search kbVBA500 kbVBA600 kbVB500 kbVB600
Version           : WINDOWS:5.0,6.0
Issue type        : kbhowto

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