如何使用两个查询的联盟，通过有秩序SQL和计数（）

Question

我有以下两个查询如何使用union，以便看到无论是在单一的查询执行

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; 

样品的结果出把

样品出来把

Answer 1

只需添加UNION ALL在这些查询之间。当UNION ALL应用时，ORDER BY子句不会接受。所以我通过把它们放在内部集合中来结束它。

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。如果您的情况是分开下单，那么您可以通过内部订单进行下单。

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 2

您可以通过row_number() over(partition by..喜欢这样做，

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 

或者你可以使用union all等;

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 3

如果您想使用联合，您可以尝试以下查询。

(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，我建议您使用where age IN (7,9)这将减少工作量。

Answer 4

当UNION语句中的任何SELECT语句包含ORDER BY子句时，该子句应放置在所有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

你必须使用类似的东西。我不使用工会，但得到2个不同年龄组的计数，然后选择其中最大计数

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，修改或删除WHERE age IN (7, 9)子句，以获得count更多age组。