我不相信Bittercode,因为他说LINQ将超越正则表达式。 所以我做了一个小测试只是为了确定。如何做到这一点
三个例子:
Dim _invalidChars As Char() = New Char() {"j"c, "a"c, "n"c}
Dim _textToStrip As String = "The quick brown fox jumps over the lazy dog"
Private Sub btnStripInvalidCharsLINQ_Click(sender As System.Object, e As System.EventArgs) Handles btnStripInvalidCharsLINQ.Click
Dim stripped As String = String.Empty
Dim sw As Stopwatch = Stopwatch.StartNew
For i As Integer = 0 To 10000
stripped = _textToStrip.Where(Function(c As Char) Not _invalidChars.Contains(c)).ToArray
Next
sw.Stop()
lblStripInvalidCharsLINQ.Text = _stripped & " - in " & sw.Elapsed.TotalMilliseconds & " ms"
End Sub
Private Sub btnStripInvalidCharsFOR_Click(sender As System.Object, e As System.EventArgs) Handles btnStripInvalidCharsFOR.Click
Dim stripped As String = String.Empty
Dim sw As Stopwatch = Stopwatch.StartNew
stripped = _textToStrip
For i As Integer = 0 To 10000
For Each c As Char In _invalidChars
stripped = stripped.Replace(c, "")
Next
Next
sw.Stop()
lblStipInvalidcharsFor.Text = stripped & " - in " & sw.Elapsed.TotalMilliseconds & " ms"
End Sub
Private Sub btnStripInvalidCharsREGEX_Click(sender As System.Object, e As System.EventArgs) Handles btnStripInvalidCharsREGEX.Click
Dim stripped As String = String.Empty
Dim sw As Stopwatch = Stopwatch.StartNew
For i As Integer = 0 To 10000
stripped = Regex.Replace(_textToStrip, "[" & New String(_invalidChars) & "]", String.Empty)
Next
sw.Stop()
lblStripInvalidCharsRegex.Text = stripped & " - in " & sw.Elapsed.TotalMilliseconds & " ms"
End Sub
结果:
所以,for循环与outperformes与string.replace所有其他方法。
因此,我会对字符串对象做一个扩展函数。
Module StringExtensions
<Extension()> _
Public Function ReplaceAll(ByVal InputValue As String, ByVal chars As Char(), replaceWith As Char) As String
Dim ret As String = String.Empty
For Each c As Char In chars
ret.Replace(c, replaceWith)
Next
Return ret
End Function
然后,你可以使用这个功能不错,可读取一行:
_textToStrip.ReplaceAll(_invalidChars, CChar(String.Empty))
来源
2012-11-21 07:55:13
JDC
这是一个“应该问精确问题而不是改写它”的情况。详细说来,我实际上有一个字符串,它基本上是一系列由空格分隔的单词。我有一串我想从字符串中删除的单词,因此我认为为什么我认为每个foreach都可能/有用。当以这种方式询问(不同的)问题时,正则表达式不适用。所以基本上: dim words()as string =(“the”,“brown”,“lazy”) dim sentence as string =“快速棕色狐狸跳跃” results =“快速狐狸跳跃” 我的希望是words.ForEach(Function(w)sentence.Replace(w,“”) – hitch 2009-08-26 06:45:41