How To Format Strings to Right-Justify When Printing

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

95945 How to Right Justify Numbers Using Format$
79094 PRB: Format$ Using # for Digit Affects Right Alignment
Propriétés

ID d'article : 217012 - Dernière mise à jour : 1 juil. 2004 - Révision : 1

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