优化联合 sql 查询

Question

我在 3 个表之间做一个 UNION。为了让您了解情况，一张表被视为主表，另外两张表被视为子表。 2 个子表对于每种情况始终具有相同数量的记录。所以我想在这3个表之间建立一个联合，在这个联合中我想重复在主表中获取的列中的信息，并列出2个子表的信息，其中子表1中的每一行都对应于该行在子表2中。由于专业原因，我无法放置真实的代码，所以这里是一个与我的真实案例相对应的简单示例。 考虑一个名为 Author (Author_ID, Author_FirstName, Author_LastName) 的表。该表将是我们的主表。然后考虑我们有一个名为 Adresses (Adress_ID, Street_Coord, Author_ID) 的表，这将是我们的子表 1。然后考虑名为 Cities 的表 (City_ID, City_Name, Author_ID)。我们的作者 X 在 2 个城市有 2 个地址。当我执行我的查询时，我得到了完全合乎逻辑的结果 R1，但我想修改我的查询以获得结果 R2。您能帮我更改查询以获得结果 R2 吗？

SQL 查询：

SELECT "Author"."Author_ID", "Author"."Author_FirstName", "Author.Author_LastName",
       TO_CHAR(NULL) AS "Street_Coord", TO_CHAR(NULL) AS "City_Name"
FROM "Author"
WHERE "Author"."Author_ID"='X'

UNION

SELECT TO_NUMBER(NULL) AS "Author_ID", TO_CHAR(NULL) AS "Author_FirstName", TO_CHAR(NULL) AS "Author_LastName", 
       "Adresses"."Street_Coord", TO_CHAR(NULL) AS "City_Name" 
FROM "Adresses"
WHERE "Adresses"."Author_ID"='X'

UNION

SELECT TO_NUMBER(NULL) AS "Author_ID", TO_CHAR(NULL) AS "Author_FirstName", TO_CHAR(NULL) AS "Author_LastName", 
       TO_CHAR(NULL) AS "Street_Coord", "Cities"."City_Name"
FROM "Cities"
WHERE "Cities"."Author_ID"='X'

结果 R1：

ID_AUTHOR | AUTHOR_FirstName | AUTHOR_LastName | Street_Coord   | City_Name       |
----------------------------------------------------------------------------------
X         |James             | Conor           | NULL           | NULL            |
----------------------------------------------------------------------------------
X         |NULL              | NULL            | 1245 rich st   | NULL            |
----------------------------------------------------------------------------------
X         |NULL              | NULL            | 154 music st   | NULL            |
----------------------------------------------------------------------------------
X         |NULL              | NULL            | NULL           | Madrid          |
----------------------------------------------------------------------------------
X         |NULL              | NULL            | NULL           | Barcelona       |
----------------------------------------------------------------------------------

结果 R2：我希望你帮助得到这个结果：

ID_AUTHOR | AUTHOR_FirstName | AUTHOR_LastName | Street_Coord   | City_Name       |
----------------------------------------------------------------------------------
X         |James             | Conor           | 1245 rich st   | Madrid          |
----------------------------------------------------------------------------------
X         |James             | Conor           | 154 music st   | Barcelona       |
----------------------------------------------------------------------------------

非常感谢， 瓦卢德

Answer 1

首先想到的是您真的想将这些表连接在一起。比如：

SELECT "Author"."Author_ID", "Author"."Author_FirstName", "Author.Author_LastName",
       "Adresses"."Street_Coord", "Cities"."City_Name"
FROM "Author" join
     "Adresses"
     on Author.Author_id = Adresses.Author_id join
     Cities
     on Author.Author_id = Cities.Author_id 
WHERE "Author"."Author_ID"='X';

但是，这将返回值的笛卡尔积，即 4 行而不是 2 行。似乎目的是“对齐”不同的表格。在这种情况下，这个查询接近你想要的：

SELECT "Author"."Author_ID", "Author"."Author_FirstName", "Author.Author_LastName",
       "Adresses"."Street_Coord", "Cities"."City_Name"
FROM "Author" join
     (select a.*, rownum as seqnum from "Adresses" a
     ) Adresses
     on Author.Author_id = Adresses.Author_id join
     (select c.*, rownum as seqnum from Cities c
     ) Cities
     on Author.Author_id = Cities.Author_id and addresses.seqnum = cities.seqnum
WHERE "Author"."Author_ID"='X';

这种方法的主要问题是 SQL 表本质上是无序的。您需要一些列来对这些进行排序，以保证查询正常工作。例如，如果您在表中有一个自动递增的 id，那么每个子查询可以是 order by id。

Answer 2

你不想做一个 UNION，你想做一个 JOIN ：

SELECT "Author"."Author_ID", "Author"."Author_FirstName", "Author.Author_LastName", "Adresses"."Street_Coord", "Cities"."City_Name"
 FROM "Author"
 INNER JOIN "Adresses" ON "Author"."Author_ID"="Adresses"."Author_ID"
 INNER JOIN "Cities" ON "Author"."Author_ID"="Cities"."Author_ID"
 WHERE "Author"."Author_ID"='X'

没有测试，可能有错别字...

此外，您需要在“地址”和“城市”之间添加一个额外的连接列，以将“1245 Rich st”与“Madrid”、“154 music st”和“Barcelona”相匹配。类似“City_ID”的东西。然后将其添加到 INNER JOIN ... ON 子句中：

INNER JOIN "Cities" ON "Adresses"."City_ID"="Cities"."City_ID" AND "Author"."Author_ID"="Cities"."Author_ID"