Привет, я могу сделать это в синтаксисе метода, но я пытаюсь улучшить свои навыки лямбды, как я могу это сделать:
SELECT SUM([job_group_quota]) as 'SUM'
FROM [dbo].[tbl_job_session]
WHERE [job_group_job_number] = @jobnum
and [job_group_ID] like @sess
GROUP BY [job_group_job_number]
Я занимаюсь этим, но не могу понять.
lnq.tbl_job_sessions.GroupBy(a => a.job_group_job_number == jnum)
.Select(b => new { b.job_group_quota}).Sum();
Общий пример:
query
.GroupBy(item => item.GroupKey)
.Select(group => group.Sum(item => item.Aggregate));
Несколько групп по примерам
public void GroupBy1() { var personList = dbEntities.People.GroupBy(m = > m.PersonType). Выберите (m = > new {PersonType = m.Key, Count = m.Count()}); }
public void GroupBy2()
{
var personList = dbEntities.People.GroupBy(m => new { m.PersonType, m.FirstName }).Select(m => new { PersonType = m.Key, Count = m.Count() });
}
public void GroupBy3()
{
var personList = dbEntities.People.Where(m => m.EmailPromotion != 0).GroupBy(m => new { m.PersonType, m.FirstName }).Select(m => new { PersonType = m.Key, Count = m.Count() });
}
public void GroupBy4()
{
var personList = dbEntities.People.GroupBy(m => new { m.PersonType, m.FirstName }).Where(m => m.Count() > 70).Select(m => new { PersonType = m.Key, Count = m.Count() });
}
public void GroupBy5()
{
var personList = dbEntities.People
.GroupBy(m =>
new
{
m.PersonType
}).Where(m => m.Count() > 70)
.Select(m =>
new
{
PersonType = m.Key,
Count = m.Count()
});
var list1 = dbEntities.People.
GroupBy(m => new { m.PersonType }).
Select(m =>
new
{
Type = m.Key,
Count = m.Count()
})
.Where(
m => m.Count > 70
&& m.Type.PersonType.Equals("EM")
|| m.Type.PersonType.Equals("GC"));
}
public void GroupBy6()
{
var list1 = dbEntities.People.
GroupBy(m => new { m.PersonType, m.EmailPromotion }).Select(m =>
new
{
Type = m.Key,
Count = m.Count()
})
.Where
(
m => m.Count > 70 && m.Type.EmailPromotion.Equals(0) &&
(
m.Type.PersonType.Equals("EM") ||
m.Type.PersonType.Equals("GC")
));
}
public void GroupBy7()
{
var list1 = dbEntities.People.
GroupBy(m => m.PersonType).
Select(c =>
new
{
Type = c.Key,
Total = c.Sum(p => p.BusinessEntityID)
});
}
public void GroupBy8()
{
var list1 = dbEntities.People.
GroupBy(m => m.PersonType).
Select(c =>
new
{
Type = c.Key,
Count = c.Count(),
Total = c.Sum(p => p.BusinessEntityID)
});
}
public void GroupBy9()
{
var list1 = dbEntities.People.
GroupBy(m => m.PersonType).
Select(c =>
new
{
Type = c.Key,
Max = c.Max(),
});
}
В этом примере показано, как итерировать сгруппированные значения, получая ключ и итоговые значения, и как получать итоговые значения напрямую (как и предыдущие). Оба используют только лямбда-оператор.
using System;
using System.Collections.Generic;
using System.Linq;
public class Person
{
public string Name { get; set; }
public string City { get; set; }
public string ZipCode { get; set; }
public int SomeValue { get; set; }
}
class Program
{
static void Main(string[] args)
{
List<Person> data = GetPopulatedData();
var totals = data.GroupBy(x =>
new { x.Name, x.City, x.ZipCode }).Select(y =>
y.Sum(i => i.SomeValue));
var groupsForIterate = data.GroupBy(x =>
new { x.Name, x.City, x.ZipCode });
Console.WriteLine("Totals: ");
foreach (var total in totals)
{
Console.WriteLine(total);
}
Console.WriteLine("Categories: ");
foreach (var categ in groupsForIterate)
{
// You can refer to one field like this: categ.Key.Ciduad
Console.WriteLine("Group" + categ.Key);
Console.WriteLine(categ.Sum(x => x.SomeValue));
}
//Output:
//Totals:
//1
//2
//1
//Categories:
//Group{ Name = Mark, City = BCN, ZipCode = 00000 }
//1
//Group{ Name = Mark, City = BCN, ZipCode = 000000 }
//2
//Group{ Name = John, City = NYC, ZipCode = 000000 }
//1
}
private static List<Person> GetPopulatedData()
{
List<Person> datos = new List<Person>()
{
new Person(){Name="Mark", City = "BCN",
ZipCode = "00000", SomeValue = 1}, // group A
new Person(){Name="Mark", City = "BCN",
ZipCode = "000000", SomeValue = 1}, // group B
new Person(){Name="Mark", City = "BCN",
ZipCode = "000000", SomeValue = 1}, // group B
new Person(){Name="John", City = "NYC",
ZipCode = "000000", SomeValue = 1}, // group C
};
return datos;
}
}
Если вы хотите получить ключевой к сумме словарный результат.
var allJobQuota = jobSessions.GroupBy(s => s.jobNumber)
.ToDictionary(g => g.Key, g => g.Sum(s => s.quota));
Sum Ficha_Venda and Entrada from Movimento:
var query = from bd in db.Movimento
where (movimento.Data != null ? bd.Data == movimento.Data : bd.Data == movimento.Data)
&& (bd.Loja == Loja)
group bd by bd.Data into t
select new {entrada = t.Sum(bd=> bd.Entrada), ficha = t.Sum(bd=> bd.Ficha_Venda)};