Microsoft KB Archive/217012

= How To Format Strings to Right-Justify When Printing =

Article ID: 217012

Article Last Modified on 7/1/2004

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APPLIES TO


 * Microsoft Visual Basic 5.0 Learning Edition
 * Microsoft Visual Basic 6.0 Learning Edition
 * Microsoft Visual Basic 5.0 Professional Edition
 * Microsoft Visual Basic 6.0 Professional Edition
 * Microsoft Visual Basic 5.0 Enterprise Edition
 * Microsoft Visual Basic 6.0 Enterprise Edition

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This article was previously published under Q217012



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.



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
 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.

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