Sql GroupBy返回超出限制的结果

我试图为分组但在where语句之外的数据包含默认值

表格

Name    Location
-----------------------
Chris   North
John    North
Jane    North-East
Bryan   South

查询

SELECT 
    Location,
    COUNT(*)
FROM Users
WHERE Location = 'North' OR Location = 'North-East'
GROUP BY Location

输出

North   2
North-East  1

North   2
North-East  1
South   0

所需输出

North   2
North-East  1

North   2
North-East  1
South   0

是否可以为where子句之外的每个位置返回零

更新

Name    Location
-----------------------
Chris   North
John    North
Jane    North-East
Bryan   South

谢谢大家的帮助。我最终使用了左连接，因为这对我来说是最快的，并且产生了正确的结果

DECLARE @Locations as Table(Name varchar(20));
DECLARE @Users as Table(Name varchar(20), Location varchar(20));

INSERT INTO @Users VALUES ('Chris', 'North')
INSERT INTO @Users VALUES ('John', 'North')
INSERT INTO @Users VALUES ('Jane', 'North-East')
INSERT INTO @Users VALUES ('Bryan', 'South')

INSERT INTO @Locations VALUES ('North')
INSERT INTO @Locations VALUES ('North-East')
INSERT INTO @Locations VALUES ('South')

SELECT 
    l.Name, 
    count(u.location)
FROM 
    @Locations l 
LEFT JOIN
     @Users u on l.Name = u.location and u.location in ('North', 'North-East')
group by 
l.Name;

---

我认为最简单的方法是使用条件聚合：

SELECT Location,
       SUM(CASE WHEN Location IN ('North', 'North-East') THEN 1 ELSE 0 END) as cnt
FROM Users u
GROUP BY Location;

或者，如果您有位置表，则更好：

select l.location, count(u.location)
from locations l left join
     users u
     on l.location = u.location and
        u.location in ('North', 'North-East')
group by l.location;

---

我认为最简单的方法是使用条件聚合：

SELECT Location,
       SUM(CASE WHEN Location IN ('North', 'North-East') THEN 1 ELSE 0 END) as cnt
FROM Users u
GROUP BY Location;

或者，如果您有位置表，则更好：

select l.location, count(u.location)
from locations l left join
     users u
     on l.location = u.location and
        u.location in ('North', 'North-East')
group by l.location;

---

假设没有locations表，唯一的方法是执行DISTINCT和sub select

SELECT DISTINCT
    Location,
    (SELECT COUNT(*) FROM users AS U 
     WHERE U.Name = Users.Name
          AND Location = 'North' OR Location = 'North-East')
FROM Users
WHERE Location = 'North' OR Location = 'North-East'

---

此代码执行大量表扫描，并且在生产环境中的大型表上运行时可能会导致系统出现问题，在生产环境中，此查询将一天运行多次。

假设没有位置表，唯一的方法是执行DISTINCT和sub select

SELECT DISTINCT
    Location,
    (SELECT COUNT(*) FROM users AS U 
     WHERE U.Name = Users.Name
          AND Location = 'North' OR Location = 'North-East')
FROM Users
WHERE Location = 'North' OR Location = 'North-East'

---

此代码执行大量表扫描，并且在生产环境中的大型表上运行此查询时，可能会导致系统出现问题。

以及基于@GordonLinoff answer的第三个选项，但给出正确的结果如下所示。这使用一个子查询来获取位置的不同列表，并将其左连接到users表

select l.location, count(u.location)
from (SELECT DISTINCT Location FROM Users) l 
left join users u
     on l.location = u.location and
        u.location in ('North', 'North-East')
group by l.location

---

第三个选项基于@GordonLinoff答案，但给出正确的结果如下。这使用一个子查询来获取位置的不同列表，并将其左连接到users表

select l.location, count(u.location)
from (SELECT DISTINCT Location FROM Users) l 
left join users u
     on l.location = u.location and
        u.location in ('North', 'North-East')
group by l.location

---

你的第二个是我要去的地方，但它给出了一个不正确的结果（4代表

北方

）你的第二个是我要去的地方，但它给出了一个不正确的结果（4代表

北方

）