我有两张桌子。
CREATE TABLE designs
( game_id INT NOT NULL,
des_id INT NOT NULL,
PRIMARY KEY(game_id, des_id),
FOREIGN KEY(game_id) REFERENCES Game(id),
ON UPDATE CASCADE)
CREATE TABLE designer
( name VARCHAR(30) NOT NULL,
id INT NOT NULL,
PRIMARY KEY(id),
FOREIGN KEY(id) REFERENCES designs(des_id),
ON UPDATE CASCADE); 让我说我有数据:
设计:
0--0
0--1
1-2
2-3
2-4
..
设计师:
鲍勃--0
吉尔-1
鲍勃--2
罗伯-3
吉尔-4
在更新之后,我希望“设计”表看起来像:
0--0
0--1
1--0
2-3
2-1
我需要什么更新查询来完成这个任务?
我尝试过的一些查询是:
UPDATE designs
SET des_id = (
SELECT a.id
FROM designer as a
JOIN designer as b
ON a.name=b.name AND a.id < b.id
WHERE des_id = b.id); ..。
UPDATE `designs` as a
JOIN designer as b
ON a.des_id=b.id
SET a.des_id = b.id
WHERE b.id = (
SELECT c.id
FROM designer as c
LEFT JOIN designer as d
ON c.name=d.name
WHERE c.id<d.id)发布于 2014-11-22 12:29:12
我有个主意。请注意,它使用了以“group/order”的形式进行的有记录的黑客攻击:
UPDATE designs d
JOIN
( select d1.id matcher_id
, d2.id select_id
from `designer` d1
JOIN designer d2
ON d1.name = d2.name
group
by d1.id
Order
by d2.id
) x
ON x.matcher_id = d.des_id
SET d.des_id = select_id发布于 2014-11-22 12:52:44
你的左加入想法几乎是正确的,但这里有另一个更快的想法.
DROP TABLE IF EXISTS designs;
CREATE TABLE designs
( game_id INT NOT NULL
, designer_id INT NOT NULL
, PRIMARY KEY(game_id, designer_id)
);
DROP TABLE IF EXISTS designers;
CREATE TABLE designers
( name VARCHAR(30) NOT NULL
, designer_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
);
INSERT INTO designs VALUES
(1,1),
(1,2),
(2,3),
(3,4),
(3,5);
INSERT INTO designers VALUES
('Bob',1),
('Jill',2),
('Bob',3),
('Rob',4),
('Jill',5);
SELECT * FROM designs;
+---------+-------------+
| game_id | designer_id |
+---------+-------------+
| 1 | 1 |
| 1 | 2 |
| 2 | 3 |
| 3 | 4 |
| 3 | 5 |
+---------+-------------+
SELECT * FROM designers;
+------+-------------+
| name | designer_id |
+------+-------------+
| Bob | 1 |
| Jill | 2 |
| Bob | 3 |
| Rob | 4 |
| Jill | 5 |
+------+-------------+
UPDATE designs g
JOIN designers d
ON d.designer_id = g.designer_id
JOIN designers x ON x.name = d.name
JOIN
( SELECT name
, MIN(designer_id) min_designer_id
FROM designers
GROUP
BY name
) y
ON y.name = x.name
AND y.min_designer_id = x.designer_id
SET g.designer_id = x.designer_id;
SELECT * FROM designs;
+---------+-------------+
| game_id | designer_id |
+---------+-------------+
| 1 | 1 |
| 1 | 2 |
| 2 | 1 |
| 3 | 2 |
| 3 | 4 |
+---------+-------------+实际上,在更新的特殊情况下,我认为这也同样有效,我不太确定这是否会降低性能.
UPDATE designs g
JOIN designers x
ON x.designer_id = g.designer_id
JOIN designers y
ON y.name = x.name
AND y.designer_id < x.designer_id
SET g.designer_id = y.designer_id;https://stackoverflow.com/questions/27077092
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