mysql - 仅加入值与变量匹配的特定行

Question

我有多个包含不同列数的 MySQL 表。加入其中三个表后，我得到了一个结构如下的结果表:

+------------+------------+-----------+-------+------+
| student_id | first_name | last_name | class | rank |
+------------+------------+-----------+-------+------+
| 1          | John       | Doe       | 2012  | 1    |
+------------+------------+-----------+-------+------+
| 2          | Suzy       | Public    | 2013  | 12   |
+------------+------------+-----------+-------+------+
| 3          | Mike       | Smith     | 2014  | 50   |
+------------+------------+-----------+-------+------+

我还有两个未参与初始连接的附加表:

兴趣

+-------------+------------+-----------------------+----------------+
| interest_id | student_id | employer_interest     | interest_level |
+-------------+------------+-----------------------+----------------+
| 1           | 1          | Wayne Enterprises     | High           |
+-------------+------------+-----------------------+----------------+
| 2           | 1          | Gotham National Bank  | Medium         |
+-------------+------------+-----------------------+----------------+
| 3           | 2          | Wayne Enterprises     | Low            |
+-------------+------------+-----------------------+----------------+
| 4           | 3          | Gotham National Bank  | High           |
+-------------+------------+-----------------------+----------------+

优惠

+----------+------------+-----------------------+
| offer_id | student_id | employer_offer        |
+----------+------------+-----------------------+
| 1        | 1          | Wayne Enterprises     |
+----------+------------+-----------------------+
| 2        | 1          | Gotham National Bank  |
+----------+------------+-----------------------+
| 3        | 2          | Wayne Enterprises     |
+----------+------------+-----------------------+

interest 和 offers 表不一定包含每个 student_id 的记录，但同时包含引用一个单个 student_id。

对于后两个表中的每一个，我想:

1. 选择 employer_interest 或 employer_offer 值等于 $var(我在 PHP 中设置的变量)的所有行<
2. 将这些行连接到原始表

例如，如果 $var 设置为 Wayne Enterprises，我希望结果表为:

+------------+------------+-----------+-------+------+-------------------+----------------+-------------------+
| student_id | first_name | last_name | class | rank | employer_interest | interest_level | employer_offer    |
+------------+------------+-----------+-------+------+-------------------+----------------+-------------------+
| 1          | John       | Doe       | 2012  | 1    | Wayne Enterprises | High           | Wayne Enterprises |
+------------+------------+-----------+-------+------+-------------------+----------------+-------------------+
| 2          | Suzy       | Public    | 2013  | 12   | Wayne Enterprises | Low            | Wayne Enterprises |
+------------+------------+-----------+-------+------+-------------------+----------------+-------------------+
| 3          | Mike       | Smith     | 2014  | 50   | NULL              | NULL           | NULL              |
+------------+------------+-----------+-------+------+-------------------+----------------+-------------------+

仅使用 MySQL 查询是否可以实现我正在尝试做的事情？如果可以，我该怎么做？

Answer 1

听起来您只需要对其他表进行 LEFT JOIN，因为您似乎希望看到第一组中的所有学生，而不管任何工作机会/兴趣如何。

如果是这样...确保“Interest”和“Offers”表都有一个索引，其中学生 ID 是单个元素索引，或者是复合索引中的第一个。

select STRAIGHT_JOIN
      ORS.Student_ID,
      ORS.First_Name,
      ORS.Last_Name,
      ORS.Class,
      ORS.Rank,
      JI.Employer_Interest,
      JI.Interest,
      OFR.Employer_Offer
   from 
      OriginalResultSet ORS

         LEFT JOIN Interest JI
            ON ORS.Student_ID = JI.Student_ID
           AND JI.Employer_Interest = YourPHPVariable

            LEFT JOIN Offers OFR
               on JI.Student_ID = OFR.Student_ID
              AND JI.Employer_Interest = OFR.Employer_Offer

为了防止“NULL”结果出现在 employer interest、interest 和 offer 中，您可以将它们包装在 Coalesce() 调用中，例如(对于左连接上的所有三列)

COALESCE( JI.Employer_Interest, " " ) Employer_Interest