VB.NET String.Split方法代码示例

' 导入命名空间
Imports System.Text.RegularExpressions

Module Example
   Public Sub Main()
      Dim expressions() As String = { "16 + 21", "31 * 3", "28 / 3",
                                      "42 - 18", "12 * 7",
                                      "2, 4, 6, 8" }

      Dim pattern As String = "(\d+)\s+([-+*/])\s+(\d+)"
      For Each expression In expressions
         For Each m As Match in Regex.Matches(expression, pattern)
            Dim value1 As Integer = Int32.Parse(m.Groups(1).Value)
            Dim value2 As Integer = Int32.Parse(m.Groups(3).Value)
            Select Case m.Groups(2).Value
               Case "+"
                  Console.WriteLine("{0} = {1}", m.Value, value1 + value2)
               Case "-"
                  Console.WriteLine("{0} = {1}", m.Value, value1 - value2)
               Case "*"
                  Console.WriteLine("{0} = {1}", m.Value, value1 * value2)
               Case "/"
                  Console.WriteLine("{0} = {1:N2}", m.Value, value1 / value2)
            End Select
         Next
      Next
   End Sub
End Module

' 导入命名空间
Imports System.Text.RegularExpressions

Module Example
   Public Sub Main()
      Dim input As String = String.Format("[This is captured{0}text.]" +
                                          "{0}{0}[{0}[This is more " +
                                          "captured text.]{0}{0}" +
                                          "[Some more captured text:" +
                                          "{0}   Option1" +
                                          "{0}   Option2][Terse text.]",
                                          vbCrLf)
      Dim pattern As String = "\[([^\[\]]+)\]"
      Dim ctr As Integer = 0
      For Each m As Match In Regex.Matches(input, pattern)
         ctr += 1
         Console.WriteLine("{0}: {1}", ctr, m.Groups(1).Value)
      Next
   End Sub
End Module

' 导入命名空间
Imports System.Text.RegularExpressions

Module Example
   Public Sub Main()
      Dim input As String = "abacus -- alabaster - * - atrium -+- " +
                            "any -*- actual - + - armoir - - alarm"
      Dim pattern As String = "\s-\s?[+*]?\s?-\s"
      Dim elements() As String = Regex.Split(input, pattern)
      For Each element In elements
         Console.WriteLine(element)
      Next
   End Sub
End Module

' 导入命名空间
Imports System.Collections.Generic

Module Example
   Public Sub Main()
      Dim value As String = "This is the first sentence in a string. " +
                            "More sentences will follow. For example, " +
                            "this is the third sentence. This is the " +
                            "fourth. And this is the fifth and final " +
                            "sentence."
      Dim sentences As New List(Of String)
      Dim position As Integer = 0
      Dim start As Integer = 0
      ' Extract sentences from the string.
      Do
         position = value.IndexOf("."c, start)
         If position >= 0 Then
            sentences.Add(value.Substring(start, position - start + 1).Trim())
            start = position + 1
         End If
      Loop While position > 0
      
      ' Display the sentences.
      For Each sentence In sentences
         Console.WriteLine(sentence)
      Next
   End Sub
End Module

' This example demonstrates the String() methods that use
' the StringSplitOptions enumeration.
Class Sample
    Public Shared Sub Main() 
        Dim s1 As String = ",ONE,,TWO,,,THREE,,"
        Dim s2 As String = "[stop]" & _
                           "ONE[stop][stop]" & _
                           "TWO[stop][stop][stop]" & _
                           "THREE[stop][stop]"
        Dim charSeparators() As Char = {","c}
        Dim stringSeparators() As String = {"[stop]"}
        Dim result() As String
        ' ------------------------------------------------------------------------------
        ' Split a string delimited by characters.
        ' ------------------------------------------------------------------------------
        Console.WriteLine("1) Split a string delimited by characters:" & vbCrLf)
        
        ' Display the original string and delimiter characters.
        Console.WriteLine("1a )The original string is ""{0}"".", s1)
        Console.WriteLine("The delimiter character is '{0}'." & vbCrLf, charSeparators(0))
        
        ' Split a string delimited by characters and return all elements.
        Console.WriteLine("1b) Split a string delimited by characters and " & _
                          "return all elements:")
        result = s1.Split(charSeparators, StringSplitOptions.None)
        Show(result)
        
        ' Split a string delimited by characters and return all non-empty elements.
        Console.WriteLine("1c) Split a string delimited by characters and " & _
                          "return all non-empty elements:")
        result = s1.Split(charSeparators, StringSplitOptions.RemoveEmptyEntries)
        Show(result)
        
        ' Split the original string into the string and empty string before the 
        ' delimiter and the remainder of the original string after the delimiter.
        Console.WriteLine("1d) Split a string delimited by characters and " & _
                          "return 2 elements:")
        result = s1.Split(charSeparators, 2, StringSplitOptions.None)
        Show(result)
        
        ' Split the original string into the string after the delimiter and the 
        ' remainder of the original string after the delimiter.
        Console.WriteLine("1e) Split a string delimited by characters and " & _
                          "return 2 non-empty elements:")
        result = s1.Split(charSeparators, 2, StringSplitOptions.RemoveEmptyEntries)
        Show(result)
        
        ' ------------------------------------------------------------------------------
        ' Split a string delimited by another string.
        ' ------------------------------------------------------------------------------
        Console.WriteLine("2) Split a string delimited by another string:" & vbCrLf)
        
        ' Display the original string and delimiter string.
        Console.WriteLine("2a) The original string is ""{0}"".", s2)
        Console.WriteLine("The delimiter string is ""{0}""." & vbCrLf, stringSeparators(0))
        
        ' Split a string delimited by another string and return all elements.
        Console.WriteLine("2b) Split a string delimited by another string and " & _
                          "return all elements:")
        result = s2.Split(stringSeparators, StringSplitOptions.None)
        Show(result)
        
        ' Split the original string at the delimiter and return all non-empty elements.
        Console.WriteLine("2c) Split a string delimited by another string and " & _
                          "return all non-empty elements:")
        result = s2.Split(stringSeparators, StringSplitOptions.RemoveEmptyEntries)
        Show(result)
        
        ' Split the original string into the empty string before the 
        ' delimiter and the remainder of the original string after the delimiter.
        Console.WriteLine("2d) Split a string delimited by another string and " & _
                          "return 2 elements:")
        result = s2.Split(stringSeparators, 2, StringSplitOptions.None)
        Show(result)
        
        ' Split the original string into the string after the delimiter and the 
        ' remainder of the original string after the delimiter.
        Console.WriteLine("2e) Split a string delimited by another string and " & _
                          "return 2 non-empty elements:")
        result = s2.Split(stringSeparators, 2, StringSplitOptions.RemoveEmptyEntries)
        Show(result)
    
    End Sub
    
    
    ' Display the array of separated strings.
    Public Shared Sub Show(ByVal entries() As String) 
        Console.WriteLine("The return value contains these {0} elements:", entries.Length)
        Dim entry As String
        For Each entry In  entries
            Console.Write("<{0}>", entry)
        Next entry
        Console.Write(vbCrLf & vbCrLf)
    
    End Sub
End Class

Dim phrase As String = "The quick brown fox"
Dim words() As String

words = phrase.Split(TryCast(Nothing, Char()), 3, 
                       StringSplitOptions.RemoveEmptyEntries)

words = phrase.Split(New Char() {}, 3,
                     StringSplitOptions.RemoveEmptyEntries)

Dim phrase As String = "The quick brown fox"
Dim words() As String

words = phrase.Split(TryCast(Nothing, String()), 3, 
                       StringSplitOptions.RemoveEmptyEntries)

words = phrase.Split(New String() {}, 3,
                     StringSplitOptions.RemoveEmptyEntries)

Module Example
   Public Sub Main()
      Dim source As String = "[stop]ONE[stop][stop]TWO[stop][stop][stop]THREE[stop][stop]"
      Dim stringSeparators() As String = {"[stop]"}
      Dim result() As String
      
      ' Display the original string and delimiter string.
      Console.WriteLine("Splitting the string:{0}   '{1}'.", vbCrLf, source)
      Console.WriteLine()
      Console.WriteLine("Using the delimiter string:{0}   '{1}'.", _
                        vbCrLf, stringSeparators(0))
      Console.WriteLine()                          
         
      ' Split a string delimited by another string and return all elements.
      result = source.Split(stringSeparators, StringSplitOptions.None)
      Console.WriteLine("Result including all elements ({0} elements):", _ 
                        result.Length)
      Console.Write("   ")
      For Each s As String In result
         Console.Write("'{0}' ", IIf(String.IsNullOrEmpty(s), "<>", s))                   
      Next
      Console.WriteLine()
      Console.WriteLine()

      ' Split delimited by another string and return all non-empty elements.
      result = source.Split(stringSeparators, _ 
                            StringSplitOptions.RemoveEmptyEntries)
      Console.WriteLine("Result including non-empty elements ({0} elements):", _ 
                        result.Length)
      Console.Write("   ")
      For Each s As String In result
         Console.Write("'{0}' ", IIf(String.IsNullOrEmpty(s), "<>", s))                   
      Next
      Console.WriteLine()
   End Sub
End Module

Module Example
   Public Sub Main()
      Dim separators() As String = {",", ".", "!", "?", ";", ":", " "}
      Dim value As String = "The handsome, energetic, young dog was playing with his smaller, more lethargic litter mate."
      Dim words() As String = value.Split(separators, StringSplitOptions.RemoveEmptyEntries)
      For Each word In words
         Console.WriteLine(word)
      Next   
   End Sub
End Module

Dim phrase As String = "The quick brown fox"
Dim words() As String

words = phrase.Split(TryCast(Nothing, String()),  
                       StringSplitOptions.RemoveEmptyEntries)

words = phrase.Split(New String() {},
                     StringSplitOptions.RemoveEmptyEntries)

Dim phrase As String = "The quick brown fox"
Dim words() As String

words = phrase.Split(TryCast(Nothing, Char()),  
                       StringSplitOptions.RemoveEmptyEntries)

words = phrase.Split(New Char() {}, 
                     StringSplitOptions.RemoveEmptyEntries)

Public Class StringSplit2
   Public Shared Sub Main()
      
      Dim delimStr As String = " ,.:"
      Dim delimiter As Char() = delimStr.ToCharArray()
      Dim words As String = "one two,three:four."
      Dim split As String() = Nothing
      
      Console.WriteLine("The delimiters are -{0}-", delimStr)
      Dim x As Integer
      For x = 1 To 5
         split = words.Split(delimiter, x)
         Console.WriteLine(ControlChars.Cr + "count = {0,2} ..............", x)
         Dim s As String
         For Each s In  split
            Console.WriteLine("-{0}-", s)
         Next s
      Next x
   End Sub 
End Class

Public Class SplitTest
    Public Shared Sub Main()
        Dim words As String = "This is a list of words, with: a bit of punctuation" + _
                              vbTab + "and a tab character."
        Dim split As String() = words.Split(New [Char]() {" "c, ","c, "."c, ":"c, CChar(vbTab) })
                
        For Each s As String In  split
            If s.Trim() <> "" Then
                Console.WriteLine(s)
            End If
        Next s
    End Sub
End Class

Module Example
   Public Sub Main()
      Dim value As String = "This is a short string."
      Dim delimiter As Char = "s"c
      Dim substrings() As String = value.Split(delimiter)
      For Each substring In substrings
         Console.WriteLine(substring)
      Next
   End Sub
End Module

Public Class Tester
    Public Shared Sub Main
        Dim quote As String = "The important thing is not to " & _
            "stop questioning. --Albert Einstein"
        Dim strArray1() As String = Split(quote, "ing")
        Dim strArray2() As String = quote.Split(CChar("ing"))
        Dim counter As Integer

        For counter = 0 To strArray1.Length - 1
            Console.WriteLine(strArray1(counter))
        Next counter


        For counter = 0 To strArray2.Length - 1
            Console.WriteLine(strArray2(counter))
        Next counter
        
    End Sub


End Class