如何搜索文本中的模式，并提取符合模式的文本部分？

Question

以下问题:我尝试从文本块中读出用户名。我通常使用.Net来解决这个问题，并尝试了LINQ和Regex，但是我找不到解决方案。

用户名的模式是'jane.doe‘(没有引号)。现在，我有以下代码序列：

Imports System.Text.RegularExpressions

Public Class Form1
    Dim arrStrSplittet As String()
    Dim strRegEx As String = "[a-z]+[.]{1}[a-z]+"
    Dim regExKriterium As Regex = New Regex(strRegEx)

    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
        arrStrSplittet = stringSplitten(TextBox1.Text)
        TextBox2.Text = testFiltern(arrStrSplittet)
    End Sub

    Function testFiltern(text As String()) As String
        Dim query = From x In text Where (regExKriterium.IsMatch(x)) Select x 
        Dim strBuild As New System.Text.StringBuilder()
        For Each y As String In query
            strBuild.AppendLine(y)
        Next
        MsgBox(strBuild.ToString())
        Return strBuild.ToString()
    End Function

    Public Function stringSplitten(text As String) As String()
        Dim arrX = Split(text, vbNewLine)
        Return arrX
    End Function

End Class

我尝试以下输入：

Type Status Name
Benutzer Mustermann, Max (Server-name\max.mustermann)
Benutzer Normalverbraucher, Otto (Server-name\otto.normalverbraucher)
Benutzer Doe, Jane (Server-name\jane.doe)
Benutzer Svensson, Kalle (Server-name\kalle.svensson)
Benutzer Borg, Joe (Server-name\joe.borg)

使用上面的代码，我得到了以下输出：

Benutzer Mustermann, Max (Server-name\max.mustermann)
Benutzer Normalverbraucher, Otto (Server-name\otto.normalverbraucher)
Benutzer Doe, Jane (Server-name\jane.doe)
Benutzer Svensson, Kalle (Server-name\kalle.svensson)
Benutzer Borg, Joe (Server-name\joe.borg)

产出应是：

max.mustermann
otto.normalverbraucher
jane.doe
kalle.svensson
joe.borg

是否可以更改LINQ中的对象x？有人有别的办法解决这个问题吗？目前，我有一个工作(但相当丑陋)的解决方案通过InStr。

我希望有人能帮我。提前感谢！米索岛

Answer 1

这是我在使用.Net、Linq和Regex时所能得到的最接近可读性的格式。

var textBook = @"Type Status Name
                 Benutzer Mustermann, Max (Server-name\max.mustermann)
                 Benutzer Normalverbraucher, Otto (Server-name\otto.normalverbraucher)
                 Benutzer Doe, Jane (Server-name\jane.doe)
                 Benutzer Svensson, Kalle (Server-name\kalle.svensson)
                 Benutzer Borg, Joe (Server-name\joe.borg)";

//split by new lines
string[] lines = textBook.Split(
new string[] { Environment.NewLine },
StringSplitOptions.None
);

//removed line without regex match of username
var removedWordWithoutServerName = lines.Where(x => Regex.IsMatch(x, "[a-z]+[.]{1}[a-z]+")).ToList();

var userNames = new List<string>();

//split by Server-name\
foreach (var serverNameAndUserName in removedWordWithoutServerName.Select(serverName => serverName.Split(@"Server-name\")))
{
    //Add only matching Regex and replace ")" with ""
    userNames.AddRange(from s in serverNameAndUserName
       where Regex.IsMatch(s, "[a-z]+[.]{1}[a-z]+")
       select s.Replace(")", ""));
}

var returnUsernames = userNames;

或简单的Linq，但可读性较低。

var textBook = @"Type Status Name
                 Benutzer Mustermann, Max (Server-name\max.mustermann)
                 Benutzer Normalverbraucher, Otto (Server-name\otto.normalverbraucher)
                 Benutzer Doe, Jane (Server-name\jane.doe)
                 Benutzer Svensson, Kalle (Server-name\kalle.svensson)
                 Benutzer Borg, Joe (Server-name\joe.borg)";
            
var userNames = new List<string>();

textBook.Split(new string[] { Environment.NewLine },StringSplitOptions.None)
         .Where(x => Regex.IsMatch(x, "[a-z]+[.]{1}[a-z]+"))
         .Select(x => x.Split(@"Server-name\")).ToList()
         .ForEach(a =>
            {
                userNames.AddRange(from s in a
                    where Regex.IsMatch(s, "[a-z]+[.]{1}[a-z]+")
                    select s.Replace(")", ""));
            });

var returnNames = userNames;

Answer 2

到目前为止，你的解决方案是完全过火的。Regex.Match将返回捕获，您可以使用()指定它们

Dim users = lines
    .Select(Function(l) Regex.Match(l, "\(.+?\\(.+?)\)"))
    .Where(Function(m) m.Success)
    .Select(Function (m) m.Groups(1).Captures(0).Value)
    .ToList()

Regex如下所示

• \( a转义打开的parenthesis
• .+?任意字符，惰性匹配(最小possible)
• \\ a转义backslash
• (开始capture
• .+? any characters
• )结束capture
• \)转义结束括号