vb.net - 在 Visual Basic 中计算单词

Question

我正在尝试实现一个程序，该程序可以在您键入时计算多行文本框中的单词。我可以让它计算单词，直到我按下“enter”键并输入一个单词。它不承认这一点。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

score 2 · Accepted Answer

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

score 1 · Accepted Answer

1

这是另一个正则表达式解决方案：

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

于 2012-02-03T01:45:28.580 回答

score 0 · Accepted Answer

为什么不改成正则表达式，整个事情都可以这样

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

score 0 · Accepted Answer

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

前

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

score 0 · Accepted Answer

这是我数单词的方法

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

score 0 · Accepted Answer

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, " ")

score 0 · Accepted Answer

如果您想在不使用 Regex 的情况下轻松计算字数

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 值将是 3（无需使用修剪或替换）

score 0 · Accepted Answer

公共函数 NumWords(Txt As String) 只要
    暗淡我只要
    如果 Len(Trim(Txt)) = 0 那么
        字数 = 0
    别的
        Txt = 替换（Txt，vbNewLine，“”）
        对于 i = 255 到 2 步 -1
            Txt = 替换（Txt，空格（i），“”）
        接下来我
        NumWords = Len(Trim(Txt)) - Len(Replace(Txt, " ", "")) + 1
    万一
结束功能

score 0 · Accepted Answer

我见过很多例子，人们计算空格并认为这些相当于单词的计数。那么这是不正确的，你不应该使用这种方法。有几种情况我可以想到这不起作用：

当空格后面没有单词时，空格数为单词数的+1。
您可以有双空格，它们也将计为 +1 字。
最重要的是，您可以让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，所以整个事情会被错误地认为是一个词而不是两个词。

现在这就是我想出的。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，那么你得到了一个单词。就那么简单！适用于连续空格、换行符和几乎任何我可以扔给它的东西。

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

vb.net - 在 Visual Basic 中计算单词

9 回答 9

Related

Reference