我的客户数据库中填充了规范化地址。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;
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`);
我们的地址列表被第三方服务清理，该服务对数据进行了规范化并指出了我们有重复的位置。这是Duplicate列的来源。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，如示例数据所示。
我显然想要删除所有重复记录，但有一些用户记录指向它们。我需要它们指向不重复的地址版本。
那么如何更新AddressID中的users以匹配非重复地址？
我能想到的唯一方法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？
当所有内容完成后，数据应该如下所示：
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)
帮助？
答案 0 :(得分：2)
用户和地址之间存在多对一关系（即多个用户可以映射到同一地址）。这对我来说有点奇怪，但我想它可能有用。多对多会更有意义，即用户可以拥有多个地址，但多个用户可以共享相同的地址。通常，单个用户具有多个地址。更新架构可能有所帮助，但我离题了。
UPDATE users
-- We only care about users mapped to duplicate addresses
JOIN addresses dupe ON (users.AddressID = dupe.AddressID AND dupe.Duplicate='Y')
-- If your normalizer thingy worked right, these will all map to non-duplicates
JOIN addresses nondupe ON (dupe.Address1 = nondupe.Address1
-- Compare to other columns if you want
AND nondupe.Duplicate = '')
-- Set to the nondupe ID
SET users.AddressID = nondupe.AddressID;
答案 1 :(得分：1)
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users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，对于您提供的示例数据，＃12 Barbara的ID在SELECT查询中为空，因为她的地址被标记为重复，而实际上它对所提供的列表是唯一的。它与地址1不匹配，如“它应该如何看”结果中所示。
修改
为了处理错误的重复标记，例如＃12 Barbara，或者其他未标记的重复标记，您可以跳过重复标记列检查，只需使用ORDER BY＆amp; LIMIT在子查询上，以便它将返回第一个匹配最低的地址ID，而不管重复的标志：
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
)