将简单的 Left Outer Join 和 group by SQL 语句转换为 Linq

Convert simple Left Outer Join and group by SQL statement into Linq

2 tables: 用户和警报

Table:用户 用户ID(整数), 全名(varchar)

Table:报警 分配给(整数), 已解决(布尔)

查询:

SELECT u.Fullname, COUNT(resolved) as Assigned, SUM(CONVERT(int,Resolved)) as Resolved, COUNT(resolved) -  SUM(CONVERT(int,Resolved)) as Unresolved
FROM Alarm i LEFT OUTER JOIN Users u on i.AssignedTo = u.UserID
GROUP BY u.Fullname

结果:

Fullname  Assigned  Resolved  Unresolved
User1     204       4         200
User2     39        9         30
User3     235       200       35
User4     1         0         1
User5     469       69        400

我这辈子都想不出如何将其变成 Linq 查询。我在使用分组功能时遇到问题。 我看了无数的例子,none 有我的 Left Outer join 和分组的组合,或者它们太复杂了,我不知道如何让它与我的一起工作。如有任何帮助,我们将不胜感激!!!

更新: 我可能不清楚我在寻找什么。我正在寻找按 AssignedTo 列分组的警报,它是一个用户标识...除此之外,我想用位于用户 table 中的 FullName 替换该用户标识。有人发布并删除了一些接近的东西,除了它给了我用户 table 中的所有用户,这不是我要找的..

更新 2: 请参阅下面的回答

试试这个:

from u in context.User
join a in context.Alarm on u.UserID equals a.AssignedTo into g1
from g2 in g1.DefaultIfEmpty()
group g2 by u.Fullname into grouped
select new { Fullname = grouped.Key, Assigned = grouped.Count(t=>t.Resolved != null), Resolved = grouped.Sum
                                    (t => int.Parse(t.Resolved)), Unresolved = (grouped.Count(t=>t.Resolved != null) - grouped.Sum
                                    (t => int.Parse(t.Resolved)))}

我想在 Linq 中这个查询不一定要使用 "Grouping" 因为 "LEFT JOIN" + "GROUP BY" 的组合将它们更改为 "INNER JOIN".

    var results =
        from u in users
        join a in alarms on u.UserID equals a.AssignedTo into ua
        select new
        {
            Fullname = u.FullName,
            Assigned = ua.Count(),
            Resolved = ua.Count(a => a.Resolved),
            Unresolved = ua.Count(a => !a.Resolved)
        };

        foreach (var r in results)
        {
            Console.WriteLine(r.Fullname + ", " + r.Assigned + ", " + r.Resolved + ", " + r.Unresolved);
        }

假设您有以下型号:

这是闹钟的型号:

public class Alarm
{
    public int id { get; set; }

    public int AssignedTo { get; set; }

    [ForeignKey("AssignedTo")] 
    public virtual User User { get; set; }

    public bool Resolved { get; set; }
}

这是用户的模型:

public class User
{
    public int UserID { get; set; }

    public string FullName { get; set; }

    public virtual ICollection<Alarm> Alarms { get; set; }

    public User()
    {
        Alarms = new HashSet<Alarm>();
    }
}

这是将保存每个用户的警报统计信息的模型:

public class UserStatistics
{
    public string FullName { get; set; }
    public int Assigned { get; set; }    
    public int Resolved { get; set; }    
    public int Unresolved { get; set; }    
}

然后您可以执行以下操作:

var query = context.Users.Select(
    user =>
        new UserStatistics
        {
            FullName = user.FullName,
            Assigned = user.Alarms.Count,
            Resolved = user.Alarms.Count(alarm => alarm.Resolved),
            Unresolved = user.Alarms.Count(alarm => !alarm.Resolved)
        });


var result = query.ToList();

顺便说一句,您还可以修改查询并删除 Unresolved = user.Alarms.Count(alarm => !alarm.Resolved),然后将 Unresolved 属性 计算为这样的 属性:

public class UserStatistics
{
    public string FullName { get; set; }
    public int Assigned { get; set; }    
    public int Resolved { get; set; }    
    public int Unresolved
    {
        get { return Assigned - Resolved; }
    }
}

这将使生成的 SQL 查询更简单。

我终于明白了。

这个:

var results = alarms.GroupBy(x => x.AssignedTo)
.Join(users, alm => alm.Key , usr => usr.UserID, (alm, usr) => new {
    Fullname = usr.FullName,AssignedNum = alm.Count(),
    Resolved = alm.Where(t=>t.resolved == true).Select(y => y.resolved).Count(), 
    Unresolved = alm.Where(t=>t.resolved == false).Select(y => y.resolved).Count() });

转载:

SELECT u.Fullname, COUNT(resolved) as Assigned, SUM(CONVERT(int,Resolved)) as Resolved, 
       COUNT(resolved) -  SUM(CONVERT(int,Resolved)) as Unresolved
FROM Alarm i LEFT OUTER JOIN Users u on i.AssignedTo = u.UserID
GROUP BY u.Fullname

结果按 AssignedTo (int) 分组,但未选择 AssignedTo。相反,全名是从加入的用户 table.

中选择的

非常感谢所有试图提供帮助的人!我从你的回答中学到了很多。

为了加分,我该如何用 SQL 语法来写我的 lamdbda 答案?