在 Visual Basic 中计算单词数

Question

我正在尝试实现一个程序，在您键入时计算多行文本框中的单词数。我可以让它计算单词，直到我按下“输入”键并输入一个单词。它不认识这一点。这是我的代码：

Private Sub TextBox1_TextChanged(ByVal sender As System.Object, _
    ByVal e As System.EventArgs) Handles TextBox1.TextChanged
  Dim str As String
  Dim i, l, words As Integer
  str = TextBox1.Text

  str = LTrim(str) 'removes blank spaces at the beginning of text
  str = RTrim(str) ' removes blank spaces at the end of text
  l = str.Length
  i = 0
  words = 0
  While (i < l)
    If str(i) = " " Then
      words = words + 1
      i = i + 1
      While str(i) = " "  ' removes more than 1  blank space
        i = i + 1
      End While
    Else
      i = i + 1
    End If

  End While

  words = words + 1 ' adds the last word

  TextBox2.Text = ("" & words)

End Sub 

Answer 1

Private Sub TextBox1_TextChanged(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles TextBox1.TextChanged

    Static rex As New System.Text.RegularExpressions.Regex("\b", System.Text.RegularExpressions.RegexOptions.Compiled Or System.Text.RegularExpressions.RegexOptions.Multiline)

    Label1.Text = (rex.Matches(TextBox1.Text).Count / 2).ToString()

End Sub

Answer 2

这是另一个正则表达式解决方案：

Dim WordCount = New Regex("\w+").Matches(s).Count

Answer 3

为什么不改成正则表达式，整个事情可以像这样

Dim words As MatchCollection = Regex.Matches(value, "\s+")
words.Count

Answer 4

您需要删除“返回”键

添加这些行：

str = Replace(str, "chr(10)", "")  'Replaces  line feed
str = Replace(str, "chr(13)", "")  'Replaces carriage return

之前

  str = LTrim(str) 'removes blank spaces at the beginning of text
  str = RTrim(str) ' removes blank spaces at the end of text 

Answer 5

这是我计算单词数的方法

    Dim count As Integer = 0
    For Each word As String In Split(txtOutput.Text, " ")
        count += 1
    Next
    MsgBox("i counted " + count.ToString + " words")

Answer 6

不要使用

str = Replace(str, "chr(10)", "")  'Replaces  line feed
str = Replace(str, "chr(13)", "")  'Replaces carriage return

更好的

str = Replace(str, vbCrLf, " ")

Answer 7

如果您想在不使用正则表达式的情况下进行简单的字数统计，

Dim s As String = " Count    the   words "

Dim iWords As Integer = 0
Dim bNonPrintable As Integer = 1

Dim len As Integer = s.Length - 1
Dim i As Integer

For i = 0 To len
    If s(i) = " "c Or s(i) = Chr(10) Or s(i) = Chr(13) Then
        If bNonPrintable = 0 Then
            iWords += 1

            bNonPrintable = i
        End If
    Else
        If bNonPrintable > 0 Then bNonPrintable = 0
    End If
Next
If bNonPrintable = 0 Then iWords += 1

在这种情况下 iWords 值将为 3（无需使用修剪或替换）

Answer 8

Public Function NumWords(Txt As String) As Long
    Dim i As Long
    If Len(Trim(Txt)) = 0 Then
        NumWords = 0
    Else
        Txt = Replace(Txt, vbNewLine, " ")
        For i = 255 To 2 Step -1
            Txt = Replace(Txt, Space(i), " ")
        Next i
        NumWords = Len(Trim(Txt)) - Len(Replace(Txt, " ", "")) + 1
    End If
End Function

Answer 9

我见过很多例子，人们计算空白并认为这些相当于计算单词数。那么这是不正确的，您不应该使用此方法。我能想到在几种情况下这不起作用：

1. 当空格后面没有单词时，空格数是单词数+1。
2. 您可以有两个空格，它们也将算作 +1 个单词。
3. 最重要的是，您可以在一个单词后跟一个换行符，然后再跟另一个单词。这里没有空格，因此整个内容将被错误地视为一个单词而不是两个单词。

现在这就是我的想法。在现有字符串的开头添加一个空格。然后用空格字符替换所有出现的换行符。然后从字符串的开头开始逐个检查字符，如果当前字符是空格而后面的字符不是，那么你就得到了一个单词。就这么简单！适用于连续空格、换行符以及几乎任何我可以使用的内容。

VB6中的实现：

Private Function countwords(cstring As String) As Long
cstring = " " & cstring
cstring = Replace(cstring, vbNewLine, " ")
countwords = 0
    For i = 1 To Len(cstring) - 1
        If Mid(cstring, i, 1) = " " And Mid(cstring, i + 1, 1) <> " " Then
            countwords = countwords + 1
        End If
    Next
End Function