Question

我有两个查询如何使用union以便在单个查询执行中查看两个结果

select TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
from Health
where age = 7
group by AGE, DIAGNOSIS_CODE_1
order by total_count DESC;

select TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
from Health
where age = 9
group by AGE, DIAGNOSIS_CODE_1
order by total_count DESC;

示例输出

Answer 1

您可以row_number() over(partition by..之类的方式执行此操作，

select
    AGE, 
    DIAGNOSIS_CODE_1, 
    total_count
from (
    select  
        AGE, 
        DIAGNOSIS_CODE_1, 
        count(DIAGNOSIS_CODE_1) as total_count,
        row_number() over (partition by AGE order by count(DIAGNOSIS_CODE_1) desc) rnk
    from Health
    where age in (7, 9)
    group by AGE, DIAGNOSIS_CODE_1
) x
where rnk = 1

或者您可以使用union all之类的;

with tmp_1 as (
    select TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
    from Health
    where age = 7
    group by AGE, DIAGNOSIS_CODE_1
    order by total_count DESC
),
tmp_2 as (
    select TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
    from Health
    where age = 9
    group by AGE, DIAGNOSIS_CODE_1
    order by total_count DESC
)
select AGE, DIAGNOSIS_CODE_1, total_count from tmp_1
union all
select AGE, DIAGNOSIS_CODE_1, total_count from tmp_2

Answer 2

只需在这些查询之间添加UNION ALL即可。应用UNION ALL时，ORDER BY子句不会接受。所以我把它们放在一个内在的集合中来结束它。

    SELECT * FROM (

    SELECT TOP 1 AGE, DIAGNOSIS_CODE_1, COUNT(DIAGNOSIS_CODE_1) AS TOTAL_COUNT
    FROM HEALTH
    WHERE AGE = 7
    GROUP BY AGE, DIAGNOSIS_CODE_1


    UNION ALL

    SELECT TOP 1 AGE, DIAGNOSIS_CODE_1, COUNT(DIAGNOSIS_CODE_1) AS TOTAL_COUNT
    FROM HEALTH
    WHERE AGE = 9
    GROUP BY AGE, DIAGNOSIS_CODE_1

    )AS A
    ORDER BY TOTAL_COUNT DESC;

根据情况你可以这样做。如果您的情况是单独下订单，那么您可以在内部集合中进行订购。

    SELECT * FROM (

    SELECT TOP 1 AGE, DIAGNOSIS_CODE_1, COUNT(DIAGNOSIS_CODE_1) AS TOTAL_COUNT
    FROM HEALTH
    WHERE AGE = 7
    GROUP BY AGE, DIAGNOSIS_CODE_1
    ORDER BY TOTAL_COUNT DESC;
    )AS B

    UNION ALL

    SELECT * FROM (
    SELECT TOP 1 AGE, DIAGNOSIS_CODE_1, COUNT(DIAGNOSIS_CODE_1) AS TOTAL_COUNT
    FROM HEALTH
    WHERE AGE = 9
    GROUP BY AGE, DIAGNOSIS_CODE_1
    ORDER BY TOTAL_COUNT DESC;
    )AS A

Answer 3

如果您想使用union，可以尝试以下查询。

(select TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
from Health
where age = 7
group by AGE, DIAGNOSIS_CODE_1 ) union
(select TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
from Health
where age = 9
group by AGE, DIAGNOSIS_CODE_1) order by total_count DESC;

对于这个特定的查询，我建议你使用where age IN (7,9)来减少工作量。

Answer 4

当UNION语句中的任何SELECT语句包含ORDER BY子句时，该子句应放在所有SELECT语句之后。Using UNION of two SELECT statements with ORDER BY

SELECT * FROM (
                      SELECT TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
                      FROM Health
                      WHERE Age = 7
                      GROUP BY AGE, DIAGNOSIS_CODE_1

                     UNION ALL    

                      SELECT TOP 1 AGE, DIAGNOSIS_CODE_1, count(DIAGNOSIS_CODE_1) as total_count
                      FRPM Health
                      WHERE Age = 9
                      GROUP BY AGE, DIAGNOSIS_CODE_1
) AS T

ORDER BY total_count DESC;

Answer 5

你必须使用这样的东西。我没有使用工会，而是计算了2个不同的年龄组，然后选择其中的最大数量

的 SQLFiddle Demo

SELECT t1.age, t1.diagnosis_code_1, t1.total_count FROM (SELECT t.*, Rank() OVER ( partition BY age ORDER BY total_count DESC) AS tc FROM (SELECT age, diagnosis_code_1, Count(diagnosis_code_1) AS total_count FROM health WHERE age IN ( 7, 9 ) GROUP BY age, diagnosis_code_1 ) t ) t1 WHERE t1.tc = 1

您可以使用相同的查询并修改或删除WHERE age IN ( 7, 9 )子句，以便为更多count组获取age。

如何在SQL中使用具有order by和count（）的两个查询的并集

5 个答案: