我有一个名为“counter”的表,它有 3 个字段:
| Field | Type | Null | Key | Default | Extra |
+--------+------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| record | datetime | YES | | NULL | |
| passed | tinyint(1) | YES | | NULL | |
+--------+------------+------+-----+---------+----------------+
我的表记录
mysql> SELECT record, passed FROM counter;
+---------------------+--------+
| record | passed |
+---------------------+--------+
| 2019-09-19 00:00:00 | 1 |
我想进行查询,以根据现场记录按周和月对结果进行分组。
我尝试过这样做,但看起来不对
SELECT DATE_FORMAT(record, '%d-%m-%Y') AS ts, COUNT(*) FROM counter WHERE record >= '2019-10-09' AND record < '2019-10-09' + INTERVAL 7 DAY GROUP BY DATE_FORMAT(record, '%d-%m-%Y');
最佳答案
带有日期提取的简单分组必须有效
select
EXTRACT(YEAR FROM record) V_YEAR,
EXTRACT(MONTH FROM record) V_MONTH,
EXTRACT(WEEK FROM record) V_WEEK,
COUNT(*)
FROM COUNTER
GROUP BY
EXTRACT(YEAR FROM record) ,
EXTRACT(MONTH FROM record) ,
EXTRACT(WEEK FROM record)
使用 DATE_FORMAT(date, format)
编辑相同的作品
select
DATE_FORMAT(record, '%Y') V_YEAR,
DATE_FORMAT(record, '%M') V_MONTH,
DATE_FORMAT(record, '%d') V_WEEK,
COUNT(*)
FROM COUNTER
GROUP BY
DATE_FORMAT(record, '%Y') ,
DATE_FORMAT(record, '%M') ,
DATE_FORMAT(record, '%d')
关于MySQL GROUP BY 按周、月查询分组结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58336376/