Question

我有一个结果表，其中列出了一组值，每个值都链接到另一个包含结果日期的表。

我使用SQL来获取所有日期（使用CASE）但是我只能检索一系列结果。

    Select 
    count(CASE
    WHEN  results.test_id IN ( SELECT id 
    FROM  `test` 
    WHERE  `posted` 
    BETWEEN  '2011-07-01 00:00:00'
    AND  '2011-07-01 23:59:59') 
    THEN results.test_id
    ELSE NULL
        END) AS "1st July"          
    from `results`
    WHERE results.window_id = 2 and results.mark > 90;

我还有另一个SQL查询，它获取所有范围，但一次只能用于一个日期。

SELECT
CASE
    when mark > 90 then '>90%'
    when mark > 80 then '>80%'
    when mark > 70 then '>70%'
END as mark_results,
COUNT(*) AS count

FROM (SELECT mark from results where window_id =2) as derived
GROUP BY mark_results
ORDER BY mark_results;

我想要的是将所有内容放在统一查询中，显示每个结果范围的相关总计。如下：

Result Range | 1st July | 2nd July | 3rd July | 4th July
>90%         |    0     |    0     |    0     |    1
>80%         |    1     |    2     |    1     |    1
>70%         |    4     |    5     |    5     |    4

以便每个范围的总数显示在其日期之下。

我认为这是可能的。

Answer 1

以下语句在FROM子句中连接结果和测试。然后它按标记范围聚合查询，每天计数：

Select (CASE when mark > 90 then '>90%'
             when mark > 80 then '>80%'
             when mark > 70 then '>70%'
       END) as mark_results,
     sum(case when posted BETWEEN  '2011-07-01 00:00:00' AND  '2011-07-01 23:59:59' then 1 else 0 end) as July01,
     sum(case when posted BETWEEN  '2011-07-02 00:00:00' AND  '2011-07-02 23:59:59' then 1 else 0 end) as July02,
     . . . 
from `results` r join
     test t
     on r.test_id = t.test_id
WHERE r.window_id = 2 and results.mark > 90
group by (CASE when mark > 90 then '>90%'
               when mark > 80 then '>80%'
               when mark > 70 then '>70%'
          END)
order by 1

只需将所需日期添加到SELECT子句中即可。

我应该补充一下。。。如果你想要所有的日期，你需要将它们放在不同的行上：

Select date(posted) as PostedDate,
      (CASE when mark > 90 then '>90%'
             when mark > 80 then '>80%'
             when mark > 70 then '>70%'
       END) as mark_results,
     count(*) as cnt
     . . . 
from `results` r join
     test t
     on r.test_id = t.test_id
WHERE r.window_id = 2 and results.mark > 90
group by date(posted),
         (CASE when mark > 90 then '>90%'
               when mark > 80 then '>80%'
               when mark > 70 then '>70%'
          END)
order by 1, 2

实际上，您可能会考虑为每个日期设置一个单独的行，并将范围作为列进行旋转。

MySQL使用大小写分组多个日期和结果

1 个答案: