[英]How to create a count of values in mysql grouped by date?
I have a table called users
with columns infected
(0 || 1 || NULL) and date_checkin
(mysql timestamp), absence
(0 || 1 || NULL).我有一个名为users
的表,其中列infected
(0 || 1 || NULL) 和date_checkin
(mysql 时间戳), absence
(0 || 1 || NULL)。
I want to create a statistic, where I can see, how many guests have the infected
value 0, 1 or Zero, same for absence
grouped by the date.我想创建一个统计数据,我可以在其中看到有多少客人的infected
值为 0、1 或零,对于按日期分组的absence
也是如此。
Currently I have the amount of guests:目前我有客人数量:
select count(`id`), date(`date_checkin`) as day
from `users`
group by day
I want my result to look like我希望我的结果看起来像
| all | infected | not infected | not tested | absence | not absence | day |
| 10 | 5 | 2 | 1 | 2 | 0 | 2021-05-26 |
| ....
Here is some sample data:以下是一些示例数据:
| id | date_checkin | inf. | abs. |
---------------------------------------------------
| 42 | 2021-05-04 23:20:11 | 0 | NULL |
| 43 | 2021-05-04 23:20:14 | 0 | NULL |
| 44 | 2021-05-04 23:20:18 | NULL | 1 |
| 45 | 2021-05-04 23:20:21 | NULL | 1 |
| 46 | 2021-05-04 23:20:24 | 1 | NULL |
| 47 | 2021-05-04 23:20:28 | 0 | NULL |
| 48 | 2021-05-04 23:20:31 | 0 | NULL |
| 49 | 2021-05-04 23:20:34 | 1 | NULL |
| 50 | 2021-05-05 07:55:15 | 0 | NULL |
| 51 | 2021-05-05 11:56:41 | NULL | 1 |
| 52 | 2021-05-05 12:55:12 | 0 | NULL |
| 53 | 2021-05-09 11:13:12 | NULL | 1 |
| 54 | 2021-05-17 17:33:52 | 0 | NULL |
| 55 | 2021-05-17 17:34:42 | 0 | NULL |
| 56 | 2021-05-18 11:12:31 | 0 | NULL |
| 57 | 2021-05-18 15:09:36 | 0 | NULL |
| 58 | 2021-05-18 15:20:59 | 0 | NULL |
| 59 | 2021-05-18 15:21:16 | 0 | NULL |
| 62 | 2021-05-18 15:25:40 | 0 | NULL |
| 63 | 2021-05-18 15:29:33 | 0 | NULL |
| 64 | 2021-05-18 16:02:02 | 0 | NULL |
| 65 | 2021-05-18 16:07:33 | 1 | NULL |
| 66 | 2021-05-18 16:07:51 | 0 | NULL |
| 67 | 2021-05-18 16:09:28 | NULL | 1 |
| 68 | 2021-05-18 16:12:12 | NULL | 1 |
| 69 | 2021-05-18 16:12:31 | NULL | 1 |
| 70 | 2021-05-18 16:12:54 | NULL | 1 |
| 71 | 2021-05-18 16:22:03 | NULL | 1 |
| 72 | 2021-05-18 16:22:26 | NULL | 1 |
| 73 | 2021-05-19 11:04:27 | NULL | 1 |
| 79 | 2021-05-19 12:27:31 | NULL | 1 |
| 80 | 2021-05-21 14:28:19 | NULL | 1 |
| 81 | 2021-05-21 14:28:30 | NULL | 1 |
| 82 | 2021-05-21 14:34:39 | 0 | NULL |
| 83 | 2021-05-21 14:35:47 | NULL | 1 |
| 84 | 2021-05-21 14:36:54 | NULL | 1 |
| 85 | 2021-05-21 14:38:26 | 0 | NULL |
| 86 | 2021-05-21 14:39:38 | 0 | NULL |
| 87 | 2021-05-21 14:39:39 | 0 | NULL |
| 88 | 2021-05-21 14:39:39 | 0 | NULL |
| 89 | 2021-05-21 14:39:42 | 0 | NULL |
You can use below query:您可以使用以下查询:
SELECT
COUNT(`id`) AS counts,
SUM(CASE WHEN infected = '1' THEN 1 ELSE 0 END) AS infected,
SUM(CASE WHEN infected = '0' THEN 1 ELSE 0 END) AS not_infected,
SUM(CASE WHEN infected IS NULL THEN 1 ELSE 0 END) AS not_tested,
SUM(CASE WHEN absence = '1' THEN 1 ELSE 0 END) AS absence,
SUM(CASE WHEN absence = '0' THEN 1 ELSE 0 END) AS not_absence,
DATE(`date_checkin`) AS selected_day
FROM
users
GROUP BY
selected_day;
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