Linq等效于分组选择

Question

From a traditional SQL sentence like this:

SELECT Id, Owner, MIN(CallTime) 
FROM traffic 
WHERE CallType = "IN" 
GROUP BY Owner;

where CallTime is a datetime field, what I want is the oldest record belonging to each Owner .

How can I achieve this with Linq?

This was my attempt (I'm using Entity Framework and context is the entity instance):

var query = context.traffic.Where(t => t.CallType == "IN");
var results = query
    .GroupBy(t => t.Owner)
    .Select(g => new { CallTime = g.Min(h => h.CallTime) });

But I need also access to Id and Owner fields whereas now I have only access to CallTime .

Answer 1

You cannot access Id in the given code because you are grouping by Owner and the Key to the group will be the Owner not the 'traffic' object.

If you group by traffic objects you need some way to tell the groupBy how to compare them properly (ie group by owner) This can be done with an IEqualityComparer

eg

private class Traffic {
    public int Id { get; set; }
    public string Owner { get; set; }
    public DateTime CallTime { get; set; }
}

private class TrafficEquaityComparer : IEqualityComparer<Traffic> {
    public bool Equals(Traffic x, Traffic y) {
            return x.Owner == y.Owner;
    }

    public int GetHashCode(Traffic obj) {
        return obj.Owner.GetHashCode();
    }
}


private static TrafficEquaityComparer TrafficEqCmp = new TrafficEquaityComparer();

private Traffic[] src = new Traffic[]{
   new Traffic{Id = 1, Owner = "A", CallTime = new DateTime(2012,1,1)},  // oldest
   new Traffic{Id = 2, Owner = "A", CallTime = new DateTime(2012,2,1)},
   new Traffic{Id = 3, Owner = "A", CallTime = new DateTime(2012,3,1)},
   new Traffic{Id = 4, Owner = "B", CallTime = new DateTime(2011,3,1)},
   new Traffic{Id = 5, Owner = "B", CallTime = new DateTime(2011,1,1)},   //oldest
   new Traffic{Id = 6, Owner = "B", CallTime = new DateTime(2011,2,1)},
};

[TestMethod]
public void GetMinCalls() {
     var results = src.GroupBy(ts => ts, TrafficEqCmp)
                        .Select(grp => {
                            var oldest = grp.OrderBy(g => g.CallTime).First();
                            return new { Id = oldest.Id, 
                                         Owner = grp.Key.Owner, 
                                         CallTime = oldest.CallTime };

                        });    }

this gives

ID : Owner : MinCallTime

1 :    A   :  (01/01/2012 00:00:00)
5 :    B   :  (01/01/2011 00:00:00)

as the results.

Answer 2

Your SQL query doesn't look valid to me: you're using Id but not grouping by it. I assume that you wanted to group by Id and Owner ?

var results = query
     .GroupBy(t => new {Id = t.Id, Owner = t.Owner})
     .Select(g => new { Id = g.Key.Id, Owner = g.Key.Owner, CallTime = g.Min(h => h.CallTime) })
     .ToList();

If you want to get the oldest (smallest) ID instead of grouping by it:

var results = query
     .GroupBy(t => t.Owner)
     .Select(g => new { Id = g.Min(x => x.Id), Owner = g.Key, CallTime = g.Min(h => h.CallTime) })
     .ToList();

Answer 3

// custQuery is an IEnumerable<IGrouping<string, Customer>>

var custQuery =  
    from cust in customers  
    group cust by cust.City into custGroup  
    where custGroup.Count() > 2  
    orderby custGroup.Key  
    select custGroup;  

In that example you select the group