我在mysql中有一张桌子:
| player1 | player2 | date | fs_1 | fs_2 |
Jack Tom 2015-03-02 10 2
Mark Riddley 2015-05-02 3 1
...我需要知道有多少aces (fs_1)球员在date_g报道的比赛前(例如10天前)做过多少次比赛。
这就是我尝试过但没有成功的地方:
选项1
SELECT
players_atp.name_p AS 'PLAYER 1',
P.name_p AS 'PLAYER 2',
DATE(date_g) AS 'DATE',
result_g AS 'RESULT',
FS_1,
FS_2,
SUM(IF(date_sub(date_g, interval 10 day)< date_g, FS_1, 0)) AS 'last 10 days'
FROM
stat_atp stat_atp
JOIN
backup3.players_atp ON ID1 = id_P
JOIN
backup3.players_atp P ON P.id_p = id2
JOIN
backup3.games_atp ON id1_g = id1 AND id2_g = id2
AND id_t_g = id_t
AND id_r_g = id_r
WHERE
date_g > '2015-01-01'
GROUP BY ID1;选项2
SELECT
players_atp.name_p AS 'PLAYER 1',
P.name_p AS 'PLAYER 2',
DATE(date_g) AS 'DATE',
result_g AS 'RESULT',
FS_1,
FS_2,
SUM(CASE WHEN date_g between date_g and date_sub(date_g, interval 10 day) then fs_1 else 0 end) AS 'last 10 days'
FROM
stat_atp stat_atp
JOIN
backup3.players_atp ON ID1 = id_P
JOIN
backup3.players_atp P ON P.id_p = id2
JOIN
backup3.games_atp ON id1_g = id1 AND id2_g = id2
AND id_t_g = id_t
AND id_r_g = id_r
WHERE
date_g > '2015-01-01'
GROUP BY ID1;我已经编辑了代码,现在更容易阅读和理解。
SELECT
id1 AS 'PLAYER 1',
id2 AS 'PLAYER 2',
DATE(date_g) AS 'DATE',
result_g AS 'RESULT',
FS_1,
FS_2,
SUM(CASE
WHEN date_g BETWEEN date_g AND DATE_SUB(date_g, INTERVAL 10 DAY) THEN fs_1
END) AS 'last 20 days' FROM
stat_atp stat_atp
JOIN
backup3.games_atp ON id1_g = id1 AND id2_g = id2
AND id_t_g = id_t
AND id_r_g = id_r GROUP BY ID1;提前鸣谢。
发布于 2017-07-03 20:55:16
也许这能帮你:
SELECT
id1,
SUM(fs_1)
FROM
stat_atp
WHERE
date_g <= DATE_SUB('2015-03-02', INTERVAL 1 DAY) AND date_g >= DATE_SUB('2015-03-02', INTERVAL 10 DAY)
AND
id1='Jack'
GROUP BY id1;请记住,RDBMS用于构造严格的数据集,这些数据集通过明确的ids (在谈论SQL时使用键)相互链接。尊重三种第一种范式比较容易。这就是为什么你应该使用键来识别你的对手本身。通过这种方式,您可以使用子查询(子集)来实现您的目标。
然后,请记住,SQL是结构化的。这是它的强项和弱点,因为您将无法使用它作为一个具有循环和条件的图灵完整编程语言。但是在任何情况下,您都可以为查询找到相同的结构。但是,您可以与SQL查询结果与另一个语言交互,并在结果集本身上使用循环和条件。那随你的便。
无论如何,您可能希望阅读有关MySQL GROUP BY子句的内容,该子句与ISO表单:https://dev.mysql.com/doc/refman/5.7/en/group-by-handling.html不同。
https://stackoverflow.com/questions/44875201
复制相似问题