Question

2个表格：用户和闹钟

表：用户 用户名（INT）全名（VARCHAR）

表：报警 AssignedTo（INT）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

对于我的生活，我无法弄清楚如何将其变成Linq查询。我在分组功能方面遇到了麻烦。我看了无数的例子，没有我的左外连接与分组的组合，或者它们是如此复杂以至于我无法弄清楚如何使它与我一起工作。在这里的任何帮助将非常感谢!!!

更新的我可能不清楚我在寻找什么。我正在寻找由AssignedTo列分组的警报，这是一个用户ID ...除了，我想用user表中的FullName替换该用户ID。有人发布并删除了一些关闭的东西，除了它给了我用户表中的所有用户，这不是我正在寻找的东西..

更新2：请参阅下面的答案

Answer 1

假设您有以下型号：

这是闹钟的模型：

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User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();

顺便说一下，您还可以修改查询并删除Unresolved = user.Alarms.Count(alarm => !alarm.Resolved)，然后将Unresolved属性设为如下所示的计算属性：

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。

Answer 2

我终于明白了。

此：

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() });

再现：

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。而是从连接的用户表中选择FullName。

非常感谢所有试图提供帮助的人！我从你的答案中学到了很多东西。

对于奖励积分，我如何用SQL语法编写我的lamdbda答案？

Answer 3

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)))}

Answer 4

我想在Linq中不一定要对此查询使用“分组”，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);
        }

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

4 个答案: