将以下SQL计数等效于LINQ计数

Question

i have some problem with LINQ, my problem is similar to the example below, I have a table named History with the following data

Method DateUsed

A      2013/02/01
A      2013/01/01
B      2013/01/01
B      2012/01/01 <--
C      2013/01/01
C      2012/02/01

what i want is to get the number or the count of methods that were used in a specific year, if i want to get the year =2013 the result would be:

Method Total 

A       2
B       1 <--
C       3

in SQL my query would be something like

Select Method, Count(DateUsed) as Total from history 
where YEAR(DateUsed) = '2013' group by Method 

however, i can't figure out how to do it in LINQ, can someone help me with the LINQ equivalent?

thanks in advance

Answer 1

You ll use a query similar as your SQL for LINQ. Group by Method and then select the method and count.

var result = History.Where(H=>H.DateUsed.Year == 2013)
                    .GroupBy(g => g.Method)
                    .Select(g => new {Method = g.Key, Count = g.Count()});

Answer 2

If you're in linq to entities, you'll need SqlFunctions .

var result = context.History
                    //linq to entities
                    .Where(m => SqlFunctions.DatePart("year", DateUsed) == 2013)
                    //linq to object
                    .Where(m => m.DateUsed.Year == 2013)
                    //common part
                    .GroupBy(m => m.Method)
                    .Select(g => new {
                        Method = g.Key,
                        Total = g.Count()
                    });

Answer 3

The other answers are correct, this one just makes use of the LINQ query syntax:

var result = 
    from m in context.History
    where m.DateUsed.Year == 2013
    /* or where SqlFunctions.DatePart( "YEAR", m.DateUsed ) == 2013 */
    group m by m.Year into g
    select new { Method = g.Key, Total = g.Count() };

I find that the query syntax is easier to use in some circumstances, and that the LINQ extension methods are easier to use in others. It helps to be familiar with both.

Note: not tested