HOWTO: Format Strings to Right-Justify When Printing

ID: Q217012

The information in this article applies to:

Microsoft Visual Basic Learning, Professional, and Enterprise Editions for Windows, versions 5.0, 6.0

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.

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

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

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

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

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

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 kbGrpVB kbDSupport
Version : WINDOWS:5.0,6.0
Platform : WINDOWS
Issue type : kbhowto