具有多个 where 子句的选择查询在一个结果表中返回多个结果行？

Question

我有一个查询，例如：

SELECT COUNT(id) AS user_list FROM tbl_users
WHERE (enrolled = 1 AND age >=15 AND age < 20)
OR (enrolled = 1 AND age >=20 AND age < 25)
OR (enrolled = 1 AND age >=25 AND age < 30)
OR (enrolled = 1 AND age >=30)

如何让它为一行和一个表中的每个 where 条件返回一个结果：

user_list
---------
   18    --(enrolled = 1 AND age >=15 AND age < 20)
   12    --(enrolled = 1 AND age >=20 AND age < 25)
   22    --(enrolled = 1 AND age >=25 AND age < 30) 
   56    --(enrolled = 1 AND age >=30)

Answer 1

如前所述，您可以使用案例并在一个查询中执行所有操作，但是您不需要使用子查询，尽管我不是它的忠实粉丝，MySQL 允许您在分组，所以以下将起作用：

SELECT  CASE WHEN (enrolled = 1 AND age >=15 AND age < 20) THEN '15-19'
            WHEN (enrolled = 1 AND age >=20 AND age < 25) THEN '20-24'
            WHEN (enrolled = 1 AND age >=25 AND age < 30) THEN '25-29'
            WHEN (enrolled = 1 AND age >=30) THEN '30+'
            ELSE 'Other'
        END AS AgeRange 
        COUNT(id) AS user_list 
FROM    tbl_users
WHERE   Enrolled = 1
AND     Age >= 15
GROUP BY AgeRange;

注意，我仍然添加了 where 子句以避免扫描冗余数据。

Example on SQL Fiddle

但是，如果您缺少数据，例如将返回一个空表，而不是这个：

AGERANGE    USER_LIST
15-19       0
20-24       0
25-29       0
30+         0

正如您所料。为了解决这个问题，我将创建一个伪表以供选择，然后加入您的数据：

SELECT  t.Name AS Age_Range,
        COUNT(u.ID) AS User_List
FROM    (   SELECT  '15-19' AS Name, 15 AS LowerBound, 20 AS UpperBound, 1 AS Enrolled
            UNION ALL
            SELECT  '20-24' AS Name, 20 AS LowerBound, 25 AS UpperBound, 1 AS Enrolled
            UNION ALL
            SELECT  '25-29' AS Name, 25 AS LowerBound, 30 AS UpperBound, 1 AS Enrolled
            UNION ALL
            SELECT  '30+' AS Name, 30 AS LowerBound, 9999999 AS UpperBound, 1 AS Enrolled
        ) t
        LEFT JOIN tbl_Users u
            ON u.Enrolled = t.enrolled
            AND u.Age >= t.LowerBound
            AND u.Age < t.UpperBound
GROUP BY t.Name;

Example on SQL Fiddle

Answer 2

试试这个方法

create table t (list1 int,list2 int,list3 int,list4 int)

insert into t
SELECT 
SUM(case when age >=15 AND age < 20 then 1 else 0 end) AS user_list1,
SUM(case when age >=20 AND age < 25 then 1 else 0 end) AS user_list2,
SUM(case when age >=25 AND age < 30 then 1 else 0 end) AS user_list3,
SUM(case when age >=15 AND age < 20 then 1 else 0 end) AS user_list4
FROM tbl_users
WHERE enrolled = 1 ;

select list1 from t union all
select list2 from t union all
select list3 from t union all
select list4 from t 

Answer 3

你可以使用联合：

SELECT COUNT(id) AS user_list
  FROM tbl_users
 WHERE (enrolled = 1 AND age >= 15 AND age < 20)
union
SELECT COUNT(id)
  FROM tbl_users
 where (enrolled = 1 AND age >= 20 AND age < 25)
union
SELECT COUNT(id)
  FROM tbl_users
 where (enrolled = 1 AND age >= 25 AND age < 30)
union
SELECT COUNT(id) FROM tbl_users where (enrolled = 1 AND age >= 30)

Answer 4

使用 4 个简单的查询而不是一个复杂的查询应该不会有性能问题

Answer 5

以下 SQL 子句可以提供帮助：

Group by、Case 和 Subqueries...查找它们并与它们一起玩，自己弄清楚应该很有趣...

您可以考虑在查询中使用另一列（例如，15 到 18 之间的大小写，然后是 1，大小写...然后是 2，...）作为 Range，标识一个范围并按此分组...

这是一个示例，说明如何将您的范围标识为子查询中的列，并在按子查询中的范围分组的列之上进行查询：

select count(*) as Count, r as Range 
    from 
        (select 
        (case 
        when enrolled = 1 AND age >=20 AND age < 25 then 'r1' 
        when enrolled = 1 AND age >=25 AND age < 30 then 'r2' 
        end) as r 
        from Text)          
    as groups 
group by r

这会导致以下结果：

范围 |数

r1 | 10

r2 | 7

其中 r1 可以是与“enrolled = 1 AND age >=20 AND age

Answer 6

这很简单

select count(1),ceil(age/5)*5 from tbl_users where enrolled=1 group by ceil(age/5)