字符串合并重复项

Question

有字符串如下：
APP001□1□20■APP002□1□30■APP001□1□30■APP001□2□30■APP002□1□50■APP002□2□30■希望得到的字符串为：APP001□1□50■APP001□2□30■APP002□1□80■APP002□2□30■上面的只是一部分，其实自己都看的眼花了，
言下之意就是若有重复(前两项一样)        前两项合并，最后一项相加；不明白的可以用String.Split()方法拆分一下string[] value = content.Split('■');//拆分成项//拆分成小项
for (int i = 0; i < value.Length - 1; i++)
{
   string SubscribeNo = value[i].Split('□')[0].Trim();//编号1
   string serialno = value[i].Split('□')[1].Trim();//编号2
   string Count = value[i].Split('□')[2].Trim();//数量
}我只能做到这里了，再次说明如果有两条或两条以上 编号1 和 编号2 相等的 希望把它和成 一条数据，合成后的数量为相同项的和!!!!!!!!!!!!!!
高手请指点！！！

Answer 1

            string str = "APP001□1□20■APP002□1□30■APP001□1□30■APP001□2□30■APP002□1□50■APP002□2□30■";
            Dictionary<string, int> dic = new Dictionary<string, int>();
            foreach (string s in str.Split(new char[] { '■' }, StringSplitOptions.RemoveEmptyEntries))
            {
                string key = s.Substring(0, s.LastIndexOf('□'));
                int value = int.Parse(s.Substring(s.LastIndexOf('□') + 1));
                if (dic.ContainsKey(key))
                    dic[key] += value;
                else
                    dic.Add(key, value);
            }
            StringBuilder sb = new StringBuilder();
            foreach (string s in dic.Keys)
                sb.AppendFormat("{0}□{1}■", s, dic[s]);
            string result = sb.ToString();
            Console.WriteLine(result);/*
输出：
APP001□1□50■APP002□1□80■APP001□2□30■APP002□2□30■
*/

Answer 2

        public static void Main(string[] args)
        {
            string zz = "APP001□1□20■APP002□1□30■APP001□1□30■APP001□2□30■APP002□1□50■APP002□2□30■";
            Dictionary<string, int> dic = new Dictionary<string, int>();            string[] value= zz.Split('■');//拆分成项
            for (int i = 0; i < value.Length-1; i++)
            {              
                string k=value[i].Substring(0,value[i].LastIndexOf('□'));
                int num=Convert.ToInt32(value[i].Substring(value[i].LastIndexOf('□')+1));
                if(!dic.ContainsKey(k))
                    dic.Add(k,num);
                else
                    dic[k]+=num;
            }
            string result=string.Empty;
           foreach(string s in dic.Keys)
               result += s + "□" + dic[s] + "■";
            Console.WriteLine(result);
        }APP001□1□50■APP002□1□80■APP001□2□30■APP002□2□30■
请按任意键继续. . .

Answer 3

定义AppInfo类：
public class AppInfo
    {
        private string _name;
        private int _id;
        private int _number;        public string Name
        {
            get { return _name; }
            set { _name = value; }
        }        public int Id
        {
            get { return _id; }
            set { _id = value; }
        }        public int Number
        {
            get { return _number; }
            set { _number = value; }
        }        public AppInfo(
            string name,
            int id,
            int number)
        {
            _name = name;
            _id = id;
            _number = number;
        }        public static bool operator ==(AppInfo left, AppInfo right)
        {
            if (object.ReferenceEquals(left, right))
            {
                return true;
            }
            else if (object.ReferenceEquals(left, null)
                || object.ReferenceEquals(right, null))
            {
                return false;
            }
            else
            {
                return (left._name == right.Name) && (left._id == right._id);
            }
        }        public static bool operator !=(AppInfo left, AppInfo right)
        {
            return !(left == right);
        }        public override bool Equals(object obj)
        {
            return this == (obj as AppInfo);
        }        public override string ToString()
        {
            return string.Format("{0}□{1}□{2}■", _name, _id, _number);
        }
    }方法：
string GetString(string oldValue)
        {
            List<AppInfo> list = new List<AppInfo>();
            string[] oldValues = oldValue.Split(
                new char[] { '■' },
                StringSplitOptions.RemoveEmptyEntries);
            foreach (string value in oldValues)
            {
                string[] values = value.Split(
                    new char[] { '□' },
                    StringSplitOptions.RemoveEmptyEntries);
                if (values.Length == 3)
                {
                    AppInfo info = new AppInfo(
                        values[0],
                        int.Parse(values[1]),
                        int.Parse(values[2]));
                    AppInfo hasInfo = list.Find(new Predicate<AppInfo>(
                        delegate(AppInfo appInfo)
                        {
                            if (appInfo == info)
                            {
                                return true;
                            }
                            else
                            {
                                return false;
                            }
                        }));
                    if (hasInfo == null)
                    {
                        list.Add(info);
                    }
                    else
                    {
                        hasInfo.Number += info.Number;
                    }
                }
            }            StringBuilder sb = new StringBuilder();
            foreach (AppInfo info in list)
            {
                sb.Append(info.ToString());
            }
            return sb.ToString();
        }
用法：
MessageBox.Show(
                GetString(
                "APP001□1□20■APP002□1□30■APP001□1□30■APP001□2□30■APP002□1□50■APP002□2□30■"));

Answer 4

Dictionary<string, int> dic = new Dictionary<string, int>();
保存数据,string为key。int为数值和，通过ContainsKey判断是否有相同 key
还可使用List<T>定义哦实体类，给实体类属性赋值，判断list<T>里是否有相同的实体
分割使用string dd = "";
string[] tempArray = Regex.Split(dd,@"■",RegexOptions.IgnoreCase);
foreach(string i in tempArray)
{
 Response.Write(i.ToString());
  Response.Wite("<br />");
}
string[] tempArray = dd.Split(new char[] { '■', '□' });