如何在 C# 中读取正则表达式捕获

Question

我想在控制台中询问用户的电话号码，根据正则表达式进行检查，然后捕获数字，以便我可以按照我想要的方式格式化它们。除了正则表达式捕获部分之外，我已经完成了所有工作。如何将捕获值放入 C# 变量中？

static void askPhoneNumber()
{
    String pattern = @"[(]?(\d{3})[)]?[ -.]?(\d{3})[ -.]?(\d{4})";

    System.Console.WriteLine("What is your phone number?");
    String phoneNumber = Console.ReadLine();

    while (!Regex.IsMatch(phoneNumber, pattern))
    {
        Console.WriteLine("Bad Input");
        phoneNumber = Console.ReadLine();
    }

    Match match = Regex.Match(phoneNumber, pattern);
    Capture capture = match.Groups.Captures;

    System.Console.WriteLine(capture[1].Value + "-" + capture[2].Value + "-" + capture[3].Value);
}

Answer 1

C# 正则表达式 API 可能非常令人困惑。有组和捕获：

• A group 表示捕获组，用于从文本中提取子字符串
• 如果组出现在量词内，则每个组可以有多个“捕获”。
层次结构是：

比赛

• 团体
  • 捕捉
（一场比赛可以有多个组，每个组可以有多个捕获）

例如：

Subject: aabcabbc
Pattern: ^(?:(a+b+)c)+$

在此示例中，只有一组：

(a+b+)

。该组位于量词内，并且匹配两次。它生成两个

捕获

：aab和

abb

：

aabcabbc
^^^ ^^^
Cap1  Cap2

当一组不在量词内部时，它仅生成一个捕获。在你的例子中，你有 3 个组，每个组捕获一次。您可以使用 

match.Groups[1].Value

、

match.Groups[2].Value

和

match.Groups[3].Value

来提取您感兴趣的 3 个子字符串，而无需求助于

capture

概念。

Answer 2

//**

的行）复制到程序中，以利用在匹配中找到的值。

public static void DisplayMatchResults(Match match)
{
    Console.WriteLine("Match has {0} captures", match.Captures.Count);

    int groupNo = 0;
    foreach (Group mm in match.Groups)
    {
        Console.WriteLine("  Group {0,2} has {1,2} captures '{2}'", groupNo, mm.Captures.Count, mm.Value);

        int captureNo = 0;
        foreach (Capture cc in mm.Captures)
        {
            Console.WriteLine("       Capture {0,2} '{1}'", captureNo, cc);
            captureNo++;
        }
        groupNo++;
    }

    groupNo = 0;
    foreach (Group mm in match.Groups)
    {
        Console.WriteLine("    match.Groups[{0}].Value == \"{1}\"", groupNo, match.Groups[groupNo].Value); //**
        groupNo++;
    }

    groupNo = 0;
    foreach (Group mm in match.Groups)
    {
        int captureNo = 0;
        foreach (Capture cc in mm.Captures)
        {
            Console.WriteLine("    match.Groups[{0}].Captures[{1}].Value == \"{2}\"", groupNo, captureNo, match.Groups[groupNo].Captures[captureNo].Value); //**
            captureNo++;
        }
        groupNo++;
    }
}

使用此方法的简单示例，给定以下输入：

Regex regex = new Regex("/([A-Za-z]+)/(\\d+)$");
String text = "some/directory/Pictures/Houses/12/apple/banana/"
            + "cherry/345/damson/elderberry/fig/678/gooseberry");
Match match = regex.Match(text);
DisplayMatchResults(match);

输出为：

Match has 1 captures
  Group  0 has  1 captures '/Houses/12'
       Capture  0 '/Houses/12'
  Group  1 has  1 captures 'Houses'
       Capture  0 'Houses'
  Group  2 has  1 captures '12'
       Capture  0 '12'
    match.Groups[0].Value == "/Houses/12"
    match.Groups[1].Value == "Houses"
    match.Groups[2].Value == "12"
    match.Groups[0].Captures[0].Value == "/Houses/12"
    match.Groups[1].Captures[0].Value == "Houses"
    match.Groups[2].Captures[0].Value == "12"

假设我们要查找上述文本中上述正则表达式的所有匹配项。然后我们可以在代码中使用

MatchCollection

，例如：

MatchCollection matches = regex.Matches(text);
for (int ii = 0; ii < matches.Count; ii++)
{
    Console.WriteLine("Match[{0}]  // of 0..{1}:", ii, matches.Count-1);
    RegexMatchDisplay.DisplayMatchResults(matches[ii]);
}

其输出是：

Match[0]  // of 0..2:
Match has 1 captures
  Group  0 has  1 captures '/Houses/12/'
       Capture  0 '/Houses/12/'
  Group  1 has  1 captures 'Houses'
       Capture  0 'Houses'
  Group  2 has  1 captures '12'
       Capture  0 '12'
    match.Groups[0].Value == "/Houses/12/"
    match.Groups[1].Value == "Houses"
    match.Groups[2].Value == "12"
    match.Groups[0].Captures[0].Value == "/Houses/12/"
    match.Groups[1].Captures[0].Value == "Houses"
    match.Groups[2].Captures[0].Value == "12"
Match[1]  // of 0..2:
Match has 1 captures
  Group  0 has  1 captures '/cherry/345/'
       Capture  0 '/cherry/345/'
  Group  1 has  1 captures 'cherry'
       Capture  0 'cherry'
  Group  2 has  1 captures '345'
       Capture  0 '345'
    match.Groups[0].Value == "/cherry/345/"
    match.Groups[1].Value == "cherry"
    match.Groups[2].Value == "345"
    match.Groups[0].Captures[0].Value == "/cherry/345/"
    match.Groups[1].Captures[0].Value == "cherry"
    match.Groups[2].Captures[0].Value == "345"
Match[2]  // of 0..2:
Match has 1 captures
  Group  0 has  1 captures '/fig/678/'
       Capture  0 '/fig/678/'
  Group  1 has  1 captures 'fig'
       Capture  0 'fig'
  Group  2 has  1 captures '678'
       Capture  0 '678'
    match.Groups[0].Value == "/fig/678/"
    match.Groups[1].Value == "fig"
    match.Groups[2].Value == "678"
    match.Groups[0].Captures[0].Value == "/fig/678/"
    match.Groups[1].Captures[0].Value == "fig"
    match.Groups[2].Captures[0].Value == "678"

因此：

    matches[1].Groups[0].Value == "/cherry/345/"
    matches[1].Groups[1].Value == "cherry"
    matches[1].Groups[2].Value == "345"
    matches[1].Groups[0].Captures[0].Value == "/cherry/345/"
    matches[1].Groups[1].Captures[0].Value == "cherry"
    matches[1].Groups[2].Captures[0].Value == "345"

对于 

matches[0]

和

matches[2]

也类似。

Answer 3