日期:2014-05-17 浏览次数:20917 次
  
只写了一个,,
          List<int> age = new List<int>();
            for (int i = 0; i < AllNames.Count; i++)
            {
                if (Oname.Contains(AllNames[i]))
                {
                    for (int j = 0; j < Oname.Count(); j++)
                    {
                        if (AllNames[i] == Oname[j])
                        {
                            age.Add(Onum[j]);
                            break;
                        }
                    }
                }
                else
                {
                    age.Add(0);
                }
            }
------解决方案--------------------
原理其实是一样的
------解决方案--------------------
void Main()
{
    string[] name={"张三","李四"};
    int[] num={10,20};
    
    string[] Oname={"张三","王五"};
    int[] Onum={20,20};
    
    string[] AllNames={"张三","李四","王五"} ;
  var a1=from n in name.Select((x,i)=>new {x,i})
            join u in num.Select((y,i)=>new{y,i})
            on n.i equals u.i
            select new {n.x,u.y};
            
  var a2=from n in Oname.Select((x,i)=>new {x,i})
            join u in Onum.Select((y,i)=>new{y,i})
            on n.i equals u.i
            select new {n.x,u.y};
  
   var query= a1.Concat(a2).GroupBy(s=>s.x).Select(s=>new {s.Key,i=s.Sum(z=>z.y)});
              
   num=(from q in query
       join n in name
       on q.Key equals n into t
       from n in t.DefaultIfEmpty()
       select n==null?0:q.i).ToArray();
   
   Onum=(from q in query
       join n in Oname
       on q.Key equals n into t
       from n in t.DefaultIfEmpty()
       select n==null?0:q.i).ToArray();
   
 
}
------解决方案--------------------
//更正一下:
   num=(from q in query
       join n in a1
       on q.Key equals n.x into t
       from n in t.DefaultIfEmpty()
       select n==null?0:n.y).ToArray();
   
   Onum=(from q in query
       join n in a2
       on q.Key equals n.x into t
       from n in t.DefaultIfEmpty()
       select n==null?0:n.y).ToArray();