该表如下所示:
date name count
2013-1-1 Jack 20
2014-3-8 Jack 3
2014-3-1 Tom 1
2014-3-1 Jack 7
2014-2-28 Mary 4
2014-2-28 Tom 5我想写一个查询,在过去的30天内输出总计数为的人
eg.
Jack 10
Tom 6
Mary 4做这件事最简单的SQL是什么?
发布于 2014-03-09 18:13:53
mysql> SELECT
-> name,
-> SUM(count) as total_count
-> FROM table_name
-> WHERE `date` >= DATE_SUB(NOW(), INTERVAL 30 day) AND `date` <= NOW()
-> GROUP BY name
-> ;
+------+-------------+
| name | total_count |
+------+-------------+
| Jack | 10 |
| Mary | 4 |
| Tom | 6 |
+------+-------------+
3 rows in set (0.02 sec)旧答案:
尝试:
SELECT
name,
SUM(count) as total_count
FROM table_name
WHERE ABS(DATEDIFF(`date` , now())) < 30
GROUP BY name它正在发挥作用:
mysql> create table table_name (`date` datetime, name char(20), count int);
Query OK, 0 rows affected (0.21 sec)
mysql> insert into table_name values
-> ('2013-1-1', 'Jack', 20),
-> ('2014-3-8', 'Jack', 3),
-> ('2014-3-1', 'Tom', 1),
-> ('2014-3-1', 'Jack', 7),
-> ('2014-2-28', 'Mary', 4),
-> ('2014-2-28', 'Tom', 5);
Query OK, 6 rows affected (0.12 sec)
Records: 6 Duplicates: 0 Warnings: 0
mysql> SELECT
-> name,
-> SUM(count) as total_count
-> FROM table_name
-> WHERE ABS(DATEDIFF(`date` , now())) < 30
-> GROUP BY name;
+------+-------------+
| name | total_count |
+------+-------------+
| Jack | 10 |
| Mary | 4 |
| Tom | 6 |
+------+-------------+
3 rows in set (0.00 sec)编辑:基于评论:
如果我想得到最高计数的前2名,那就是杰克和汤姆。怎么做呢?
使用DESC命令和限制,用结果检查以下查询。
mysql> SELECT
-> name,
-> SUM(count) as total_count
-> FROM table_name
-> WHERE `date` > DATE_SUB(NOW(), INTERVAL 30 day) and `date` < NOW()
-> GROUP BY name
-> ORDER BY total_count DESC
-> LIMIT 2;
+------+-------------+
| name | total_count |
+------+-------------+
| Jack | 10 |
| Tom | 6 |
+------+-------------+
2 rows in set (0.03 sec)发布于 2014-03-09 18:20:08
你可以用:
SELECT name, SUM(count) FROM table_name WHERE date >= ( CURDATE() - INTERVAL 30 DAY ) GROUP BY name; https://stackoverflow.com/questions/22285980
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