正则表达式

[zhan1219]

在C#中如何用正则提取"@RA@YA@L2@Q2@EA"中的@前面第一个值和第二个值,如
提取后变成RYLQE和AA22A

[huming_h]

有个简单的办法，
首先置换掉所有@符号
然后根据string的index来取需要的字符，基数和偶数的组合

[huming_h]

static void Main(string[] args)
{
string s="@RA@YA@L2@Q2@EA".Replace("@","");
string s1 = "",s2="";
for (int i = 0; i < s.Length; i++)
{
if (i % 2 == 0)
{
s1 += s[i];
}
else
{
s2 += s[i];
}
}
Console.WriteLine(s1);
Console.WriteLine(s2);
Console.Read();
}

[jiangshun]

            string html = @"@RA@YA@L2@Q2@EA";
            Regex reg = new Regex(@"(?<=\@(?<s1>\w{1}))\w{1}");
            MatchCollection mc = reg.Matches(html);
            Console.WriteLine("/*\n------输出结果------------");
            string s1 = "";
            string s2 = "";
            foreach (Match m in mc)
            {
                s1 += m.Groups["s1"].ToString();
                s2 += m.Groups[0].ToString();
            }

            Console.WriteLine(s1+"\n"+s2);
            
            
            Console.WriteLine("*/");

            /*
            ------输出结果------------
            RYLQE
            AA22A
            */

[xzjxylophone]

up 学习了

[wuyi8808]

using System;
using System.Text.RegularExpressions;

class Program
{
  static void Main()
  {
    string s = "@RA@YA@L2@Q2@EA";
    string s1 = Regex.Replace(s, @"@(.).", "$1");
    string s2 = Regex.Replace(s, @"@.(.)", "$1");
    Console.WriteLine(s1);  // 输出：RYLQE
    Console.WriteLine(s2);  // 输出：AA22A
  }
}

[zpx3243]

Dim html As String = "@RA@YA@L2@Q2@EA"

Dim reg As Regex = New Regex("@(?<first>.{1})(?<second>.{1})")
Dim s1 As String = reg.Replace(html,"${first}")
Dim s2 As String = reg.Replace(html,"${second}")

<first>.{1}:红色表示匹配字符的个数，可以根据需要修改。

[yankunlin]

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