一个用c#处理字符串的问题，貌似很简单但是我想了半天还是没想出来比较好的办法。。。

[owenxin26]

假如说现在有一个字符串:
{someone}'s wallet was stolen by {someone else} but the police who called {anyone} caught him.

然后还有一个字符集合：List<string> people = new List<string> { "Tom", "Jack", "Charlie"};


要做的就是创建一个函数，把字符串中用{xxx}包围的地方替换成List里面的字符。
public string ReplaceItems(string str, List<string> items)
{
//....
}

输出结果为：
Tom's wallet was stolen by Jack but the police who called Charlie caught him.



字符串中{}数量不一定是3个，但{}的数量和对应的List集合size是相同的，{xxx}中的字符也不固定。

[Tim_Wu6]

Replace 函数

[wuyq11]

Regex reg = new Regex(@"(?<=\{)[^{}]+(?=\})");
reg.Replace(input, "");

[孟子E章]

string a = "{someone}'s wallet was stolen by {someone else} but the police who called {anyone} caught him.";
List<string> people = new List<string> { "Tom", "Jack", "Charlie"};
a = a.Replace("{someone}",people[0]).Replace("{someone else}",people[1]).Replace("{anyone}",people[2]);
Response.Write(a);

[owenxin26]

[Quote=引用 3 楼 net_lover 的回复:]

string a = "{someone}'s wallet was stolen by {someone else} but the police who called {anyone} caught him.";
List<string> people = new List<string> { "Tom", "Jack", "Charlie"};
a =……
[/Quote]


谢谢你的回答，这个方法对我举的例子是可行的，但问题是{xxx}中的字符并不一定就是{someone}，{anyone}，也可能是{abc}, {def}等其它的，所以并不适用于所有情况。

[liysky]

将字符串转换成数组，然后遍历一次，从发现{开始将要替换的对象往stringbuilder里装，直到发现第一个}。需要设置一个flag来检查{ }是否成对出现。如果不成对出现或先遇到}都要返回提示信息。即可实现

[liysky]

当然发现{前的字符也需要往stringbuilder里加

[jointan]

        public string ReplaceItems(string str, List<string> items)
        {
            int l = 0, r = 0;
            int n = items.Count;
            string temp = "";
            for (int i = 0; i < n; i++)
            {
                l = str.IndexOf("{", r);
                temp += str.Substring(r, l - r);
                temp += items[i];
                r = str.IndexOf("}", l) + 1;
            }
            temp += str.Substring(r, str.Length - r);
            return temp;
        }

[兔子-顾问]

给你写了个例子。

using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;

class Program
{
    static void Main(string[] args)
    {
        string s = "{someone}'s wallet was stolen by {someone else} but the police who called {anyone} caught him.";
        Dictionary<string, string> rule = new Dictionary<string, string>();
        rule.Add("someone", "Tom");
        rule.Add("someone else", "Jack");
        rule.Add("anyone", "Charlie");
        string result = RuleReplace(s, rule);
        Console.WriteLine(result);
        Console.ReadKey();
    }

    /// <summary>
    /// 特定的字符串替换方法
    /// </summary>
    /// <param name="src">原字符串</param>
    /// <param name="rule">替换规则</param>
    /// <returns>替换后的结果字符串</returns>
    public static string RuleReplace(string src, Dictionary<string, string> rule)
    {
        return Regex.Replace(src, @"\{([^{}]+)\}", delegate(Match m)
        {
            if (rule.ContainsKey(m.Groups[1].Value))
            {
                return rule[m.Groups[1].Value];//返回替换掉的字符串
            }
            return m.Value;//原样返回
        });
    }
}

[兔子-顾问]

你只需要自己配置rule这个字典就可以了。

[wuyi8808]

  public string ReplaceItems(string str, List<string> items)
  {
    int k = 0;
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < str.Length; i++)
    {
      if (str[i] == '{')
      {
        while (i < str.Length && str[i] != '}') i++;
        sb.Append(items[k++]);
      }
      else if (str[i] != '}') sb.Append(str[i]);
    }
    return sb.ToString();
  }

[yyd021]

   List<string> people = new List<string> { "Tom", "Jack", "Charlie"};
   int i= 0;
   while (a.Contains("{"))  //循环替换所有的“{”及内部内容。
            {
                
                string oldStr = strTemp.Substring(a.IndexOf("{"), a.IndexOf("}") - a.IndexOf("{") + 1);  //获取第一个 {xxx}

                a= a.Replace(oldStr, people[i]);
                i++;
                }

[owenxin26]

谢谢各位的回答，我最后用的方法：

public string ReplaceItems(string str, List<string> items)
{
    string textToReplace = str;
    Regex reg = new Regex(@"\{([^{}]+)\}");

    foeach(string item in items)
    {
        textToReplace = reg.Replace(textToReplace, item, 1);
    }

    return textToReplace;
}