Question

给出以下MySQL表：

| id | category | Hour | quantity|
|  0 | Sunday   |  10  |  32    |
|  0 | Sunday   |  11  |  19    |
|  0 | Sunday   |  12  |  48    |
|  0 | Sunday   |  19  |  7     |
|  1 | Monday   |  09  |  45    |
|  1 | Monday   |  10  |  17    |
|  1 | Monday   |  12  |  18    |
|  2 | Tuesday  |  08  |  16    |
|  2 | Tuesday  |  09  |  39    |
|  2 | Tuesday  |  10  |  24    |
|  2 | Tuesday  |  11  |  37    |
|  2 | Tuesday  |  12  |  40    |

我需要计算第五列，它必须是＆＃34;数量＆＃34;通过id的行数：对于0，有4行，对于1 3行，对于2 5行。

| id | category | Hour | quantity| avg |
|  0 | Sunday   |  10  |  32    |  8   |
|  0 | Sunday   |  11  |  19    | 4.75 | 
|  0 | Sunday   |  12  |  48    |  12  |
|  0 | Sunday   |  19  |  7     | 1.75 |
|  1 | Monday   |  09  |  45    |  15  |
|  1 | Monday   |  10  |  17    |  5.7 |
|  1 | Monday   |  12  |  18    |   6  |
|  2 | Tuesday  |  08  |  16    | 3.2  |
|  2 | Tuesday  |  09  |  39    | 7.8  |
|  2 | Tuesday  |  10  |  24    | 4.8  |
|  2 | Tuesday  |  11  |  37    | 7.4  |
|  2 | Tuesday  |  12  |  40    |  8   |

如何在MySQL查询中获得结果？

第一个表是此查询的结果：

 select id, category, Hour, count(*) as quantity
 FROM table_1
 GROUP by id, Hour ORDER by id, Hour;

这是我试过的，为了得到每个id出现的行数，但是我得到一个大数字，在前一个查询中id = 0次出现而不是id = 0行的计数：

select  id, Hour,  count(id)  as q
FROM table_1 
GROUP by id

这是mySql 5.6。

Answer 1

这真的非常难看和繁琐，但这是获得结果而没有主键可以使用的唯一方法：

SELECT 
  t.id, 
  t.category, 
  t.hour, 
  quantity,  
  ROUND(quantity/count,2) AS avg
FROM table_1 t
    JOIN (SELECT 
              id, Hour, count(*) as quantity
             FROM table_1
            GROUP by id, category, Hour) AS qty
    ON t.id = qty.id AND t.hour = qty.hour

    JOIN (SELECT
            id, count(distinct hour) as count 
             FROM table_1
                GROUP BY id) as counts 
    ON t.id = counts.id
GROUP BY id, hour;

至少在猜测原始数据集的样子时，它似乎在本地工作。

然而，可能有一种更简单的方式。

编辑：第二次检查时，'数量'子查询实际上并没有增加太多我能看到的内容，因此可以用'count（*）'替换，使其更加优化查询：

SELECT 
  t.id, 
  t.category, 
  t.hour, 
  count(*) as quantity,  
  ROUND(count(*)/count,2) AS avg
FROM table_1 t
    JOIN (SELECT
            id, count(distinct hour) as count 
             FROM table_1
                GROUP BY id) as counts 
    ON t.id = counts.id
GROUP BY id, hour;

Answer 2

您需要在仅按id分组的子查询中进行计数。将子查询加入主查询并进行除法。

SELECT id, category, hour, COUNT(*) AS quantity, COUNT(*)/count AS avg
FROM table_1
JOIN (SELECT id, COUNT(DISTINCT hour) AS count
      FROM table_1
      GROUP BY id) AS counts 
    ON table_1.id = counts.id
GROUP BY table_1.id, table_1.hour
ORDER BY table_1.id, table_1.hour

如何计算每个不同组中的行数？

2 个答案: