pba*_*ney 6 php mysql sql normalization duplicate-removal
我有一个填充了规范化地址的客户数据库.有重复.
每个用户都创建了自己的记录,并输入了自己的地址.因此,我们在用户和地址之间建立了一对一的关系:
CREATE TABLE `users` (
`UserID` INT UNSIGNED NOT NULL AUTO_INCREMENT,
`Name` VARCHAR(63),
`Email` VARCHAR(63),
`AddressID` INT UNSIGNED,
PRIMARY KEY (`UserID`) USING BTREE
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE `addresses` (
`AddressID` INT UNSIGNED NOT NULL AUTO_INCREMENT,
`Duplicate` VARCHAR(1),
`Address1` VARCHAR(63) DEFAULT NULL,
`Address2` VARCHAR(63) DEFAULT NULL,
`City` VARCHAR(63) DEFAULT NULL,
`State` VARCHAR(2) DEFAULT NULL,
`ZIP` VARCHAR(10) DEFAULT NULL,
PRIMARY KEY (`AddressID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
而数据:
INSERT INTO `users` VALUES
(1, 'Michael', 'michael@email.com', 1),
(2, 'Steve', 'steve@email.com', 2),
(3, 'Judy', 'judy@email.com', 3),
(4, 'Kathy', 'kathy@email.com', 4),
(5, 'Mark', 'mark@email.com', 5),
(6, 'Robert', 'robert@email.com', 6),
(7, 'Susan', 'susan@email.com', 7),
(8, 'Paul', 'paul@email.com', 8),
(9, 'Patrick', 'patrick@email.com', 9),
(10, 'Mary', 'mary@email.com', 10),
(11, 'James', 'james@email.com', 11),
(12, 'Barbara', 'barbara@email.com', 12),
(13, 'Peter', 'peter@email.com', 13);
INSERT INTO `addresses` VALUES
(1, '', '1234 Main Street', '', 'Springfield', 'KS', '54321'),
(2, 'Y', '1234 Main Street', '', 'Springfield', 'KS', '54321'),
(3, 'Y', '1234 Main Street', '', 'Springfield', 'KS', '54321'),
(4, '', '5678 Sycamore Lane', '', 'Upstate', 'NY', '50000'),
(5, '', '1000 State Street', 'Apt C', 'Sunnydale', 'OH', '54321'),
(6, 'Y', '1234 Main Street', '', 'Springfield', 'KS', '54321'),
(7, 'Y', '1000 State Street', 'Apt C', 'Sunnydale', 'OH', '54321'),
(8, 'Y', '1234 Main Street', '', 'Springfield', 'KS', '54321'),
(9, '', '1000 State Street', 'Apt A', 'Sunnydale', 'OH', '54321'),
(10, 'Y', '1234 Main Street', '', 'Springfield', 'KS', '54321'),
(11, 'Y', '5678 Sycamore Lane', '', 'Upstate', 'NY', '50000'),
(12, 'Y', '1000 Main Street', 'Apt A', 'Sunnydale', 'OH', '54321'),
(13, '', '9999 Valleyview', '', 'Springfield', 'KS', '54321');
哦,是的,让我加上那个外键关系:
ALTER TABLE `users` ADD CONSTRAINT `AddressID`
FOREIGN KEY `AddressID` (`AddressID`)
REFERENCES `addresses` (`AddressID`);
我们通过第三方服务清理了我们的地址列表,该服务对数据进行了规范化并指出了我们重复的位置.这是Duplicate专栏的来源.如果有'Y',则它是另一个地址的副本.主要地址未标记为重复,如示例数据中所示.
我显然想要删除所有重复记录,但有用户记录指向它们.我需要它们指向不重复的地址版本.
那么如何更新AddressIDin users以匹配非重复地址？
我能想到的唯一方法就是使用高级语言迭代所有数据,但我很确定MySQL拥有以更好的方式做这样的事情所需的所有工具.
这是我尝试过的:
SELECT COUNT(*) as cnt, GROUP_CONCAT(AddressID ORDER BY AddressID) AS ids
FROM addresses
GROUP BY Address1, Address2, City, State, ZIP
HAVING cnt > 1;
+-----+--------------+
| cnt | ids |
+-----+--------------+
| 2 | 5,7 |
| 6 | 1,2,3,6,8,10 |
| 2 | 4,11 |
+-----+--------------+
3 rows in set (0.00 sec)
从那里,我可以遍历每个结果行并执行此操作:
UPDATE `users` SET `AddressID` = 1 WHERE `AddressID` IN (2,3,6,8,10);
但是必须有一个更好的MySQL方式,不应该吗？
一切都说完了,数据应该是这样的:
SELECT * FROM `users`;
+--------+---------+-------------------+-----------+
| UserID | Name | Email | AddressID |
+--------+---------+-------------------+-----------+
| 1 | Michael | michael@email.com | 1 |
| 2 | Steve | steve@email.com | 1 |
| 3 | Judy | judy@email.com | 1 |
| 4 | Kathy | kathy@email.com | 4 |
| 5 | Mark | mark@email.com | 5 |
| 6 | Robert | robert@email.com | 1 |
| 7 | Susan | susan@email.com | 5 |
| 8 | Paul | paul@email.com | 1 |
| 9 | Patrick | patrick@email.com | 9 |
| 10 | Mary | mary@email.com | 1 |
| 11 | James | james@email.com | 4 |
| 12 | Barbara | barbara@email.com | 1 |
| 13 | Peter | peter@email.com | 13 |
+--------+---------+-------------------+-----------+
13 rows in set (0.00 sec)
SELECT * FROM `addresses`;
+-----------+-----------+--------------------+----------+-------------+-------+-------+
| AddressID | Duplicate | Address1 | Address2 | City | State | ZIP |
+-----------+-----------+--------------------+----------+-------------+-------+-------+
| 1 | | 1234 Main Street | | Springfield | KS | 54321 |
| 4 | | 5678 Sycamore Lane | | Upstate | NY | 50000 |
| 5 | | 1000 State Street | Apt C | Sunnydale | OH | 54321 |
| 9 | | 1000 State Street | Apt A | Sunnydale | OH | 54321 |
| 13 | | 9999 Valleyview | | Springfield | KS | 54321 |
+-----------+-----------+--------------------+----------+-------------+-------+-------+
5 rows in set (0.00 sec)
救命？
要选择您想要查看的结果：
SELECT a.UserID
,a.Name
,a.Email
,(
SELECT addressID
FROM addresses c
WHERE c.Address1 = b.Address1
AND c.Address2 = b.Address2
AND c.City = b.City
AND c.State = b.State
AND c.ZIP = b.ZIP
AND DUPLICATE != 'Y'
) as AddressID
FROM users a
JOIN addresses b
ON a.AddressID = b.AddressID
这会将用户表更新为上面查询中显示的结果。
UPDATE users a
JOIN addresses b
ON a.AddressID = b.AddressID
SET a.addressID =
(
SELECT addressID
FROM addresses c
WHERE c.Address1 = b.Address1
AND c.Address2 = b.Address2
AND c.City = b.City
AND c.State = b.State
AND c.ZIP = b.ZIP
AND Duplicate != 'Y'
)
WHERE Duplicate = 'Y'
请注意，根据您提供的示例数据，#12 Barbara 的 ID 在查询中为空SELECT，因为她的地址被标记为重复，而实际上它对于所提供的列表是唯一的。它与“它应该是什么样子”结果中所示的地址 1 不匹配。
编辑
为了处理不正确的重复标志，例如 #12 Barbara，或者可能其他未标记为这样的遗漏重复标志，您可以跳过重复标志列检查，只在子查询上使用ORDER BY& ，以便它将返回第一个最低匹配LIMIT地址 ID，无论重复标志如何：
UPDATE users a
JOIN addresses b
ON a.AddressID = b.AddressID
SET a.addressID =
(
SELECT addressID
FROM addresses c
WHERE c.Address1 = b.Address1
AND c.Address2 = b.Address2
AND c.City = b.City
AND c.State = b.State
AND c.ZIP = b.ZIP
ORDER BY c.addressID ASC
LIMIT 1
)