VB中的StringAlignment.Justify？

Question

我一直在尝试用VB打印一些文档，我一直做得很好……直到我意识到我需要用Justify打印我的文档。

这是我在PrintDocuments中使用的代码：

Dim drawcenter As New StringFormat
drawcenter.Alignment = StringAlignment.Center

e.Graphics.DrawString("Terms and Conditions of Service", New Font("Times New Roman", 13, FontStyle.Bold Or FontStyle.Underline), New SolidBrush(Color.Black), New RectangleF(50, 50, 700, 30), drawcenter)

有没有办法用Justify或者别的什么来代替StringAlignment.Center呢？

Answer 1

用于对齐左长句(或段落)的函数为你工作！

   1.  Public  Function  GetJustifiedTextinLeft (text As String, width As Integer) As String
    Dim palabras As String() = text.Split(" "c)  'text-->palabras
    Dim sb1 As New System.Text.StringBuilder()
    Dim sb2 As New System.Text.StringBuilder()
    Dim length As Integer = palabras.Length     'palabras
    Dim resultado As New System.Collections.Generic.List(Of String)()
    Dim i As Integer = 0
    While i < length
        sb1.AppendFormat("{0} ", palabras(i))   'palabras
        If sb1.ToString().Length > width Then
            resultado.Add(sb2.ToString())
            sb1 = New System.Text.StringBuilder()
            sb2 = New System.Text.StringBuilder()
            System.Math.Max(System.Threading.Interlocked.Decrement(i), i + 1)
        Else
            sb2.AppendFormat("{0} ", palabras(i))
        End If
        System.Math.Max(System.Threading.Interlocked.Increment(i), i - 1)
    End While
    resultado.Add(sb2.ToString())               'resultado

    Dim resultado2 As New System.Collections.Generic.List(Of String)()
    Dim temp As String

    Dim index1 As Integer, index2 As Integer, salto As Integer
    Dim target As String
    Dim limite As Integer = resultado.Count     'resultado
    For Each item As String In resultado        'resultado
        target = " "
        temp = item.ToString().Trim()
        index1 = 0
        index2 = 0
        salto = 2

        If limite <= 1 Then
            resultado2.Add(temp)
            Exit For
        End If
        While temp.Length <= width
            If temp.IndexOf(target, index2) < 0 Then
                index1 = 0
                index2 = 0
                target = target + " "
                System.Math.Max(System.Threading.Interlocked.Increment(salto), salto - 1)
            End If
            index1 = temp.IndexOf(target, index2)
            temp = temp.Insert(temp.IndexOf(target, index2), " ")

            index2 = index1 + salto
        End While
        System.Math.Max(System.Threading.Interlocked.Decrement(limite), limite + 1)
        resultado2.Add(temp)
    Next

    Dim builder As New System.Text.StringBuilder()
    For Each item As String In resultado2
        builder.Append(item).Append(chr(10))
   Next
    Return builder.ToString()
End Function


  2.    For Calling  above Function
  Dim  Resulttext as string=GetJustifiedTextinLeft (“YourString”,70) 
  '70 is the width 

Answer 2

你的意思是把绳子垫起来吗？

你可以试试String.PadRight。

http://www.dotnetperls.com/padright

http://msdn.microsoft.com/en-us/library/66f6d830%28v=vs.110%29.aspx

http://www.csharp-examples.net/align-string-with-spaces/

如果你想在每个单词之间有相同数量的空格，那么第一个单词在左边，最后一个单词在右边，你可能需要右转你自己的代码。

我写了这段代码，它做了我认为你想要的事情，它可以被重构：

Sub Main()

    System.Console.WriteLine("123456789012345678901234567890")
    Dim Words As New List(Of String)
    Words.Add("Hello")
    Words.Add("World")
    System.Console.WriteLine(SplitWordsOverSpace(30, Words))

    Words.Add("Cheese")
    System.Console.WriteLine(SplitWordsOverSpace(30, Words))

    Words.Add("a")
    System.Console.WriteLine(SplitWordsOverSpace(30, Words))

    System.Console.ReadLine()
End Sub

Public Function SplitWordsOverSpace(ByVal LineWidthCharCount As Integer, ByVal WordsonLine As List(Of String)) As String
    Try
        Dim TotalWordLength As Integer = 0
        For Each S As String In WordsonLine
            TotalWordLength += S.Length
        Next
        Dim LeftOverSpace As Integer = LineWidthCharCount - TotalWordLength

        Dim Spaces(WordsonLine.Count - 1) As String
        Dim SpaceperWord As Integer = Math.Floor(LeftOverSpace / (WordsonLine.Count - 1))
        Dim Remainder As Integer = LeftOverSpace Mod (WordsonLine.Count - 1)

        Dim sb As New Text.StringBuilder()
        For Each s As String In WordsonLine
            sb.Append(s)
            If Not String.Equals(s, WordsonLine(WordsonLine.Count - 1)) Then
                For i As Integer = 1 To SpaceperWord
                    sb.Append(" ")
                Next
                If Remainder > 0 Then
                    sb.Append(" ")
                    Remainder -= 1
                End If
            End If
        Next
        Return sb.ToString()
    Catch ex As Exception
        Throw ''Or something
    End Try
End Function