mysql - 根据特定值对记录进行排名的 SQL 语句

Question

我目前有以下记录:

+----+-------------+-----------------+--------+-----------------+-------------+
| id | postal_code | program_type_id | gender | school_location | school_type |
+----+-------------+-----------------+--------+-----------------+-------------+
|  1 | 66202       |               2 | female |                 |             |
|  2 | 67487       |               2 | male   | rural           | public      |
|  3 | 68504       |               2 | female | rural           | private     |
|  4 | 67554       |               2 | female | rural           | public      |
|  5 | 67212       |               2 | female | urban           | public      |
+----+-------------+-----------------+--------+-----------------+-------------+

我还有以下记录:

mysql> select id, postal_code, program_type_id, gender, school_location, school_type from applications limit 1 offset 6;
+----+-------------+-----------------+--------+-----------------+-------------+
| id | postal_code | program_type_id | gender | school_location | school_type |
+----+-------------+-----------------+--------+-----------------+-------------+
|  7 | 66202       |               2 | female | urban           | public      |
+----+-------------+-----------------+--------+-----------------+-------------+

我必须以某种方式将这条记录7与数据库中的记录进行匹配并给出分数。

评分:

匹配的 postal_code = 1000 点匹配的program_type_id = 490点匹配性别 = 20 分匹配学校类型 = 500 分

现在，我应该检索的记录应按以下顺序排列:

+----+-------------+-----------------+--------+-----------------+-------------+
| id | postal_code | program_type_id | gender | school_location | school_type |
+----+-------------+-----------------+--------+-----------------+-------------+
|  1 | 66202       |               2 | male   |                 |             | 1K points
|  3 | 68504       |               2 | female | rural           | private     | 520 points
|  2 | 67487       |               1 | male   | rural           | public      | 490 points
|  4 | 67554       |               1 | female | rural           | public      | 20 points
|  5 | 67212       |               1 | female | urban           | public      | 20 points
+----+-------------+-----------------+--------+-----------------+-------------+
5 rows in set (0.00 sec)

请注意，3 超过了 2，因为匹配 program_type_id 和性别将获得 520 分，而仅匹配 school_type 将仅获得 500 分。在本例中，3 的分数高于 2 的分数。

现在，我的问题是，有谁知道如何做到这一点以及如何做到这一点？顺便说一句，这是 MySQL 5。

Answer 1

查看 mysql 中的 case 语句。您的构造将采用以下形式

select <fields that matter>, (case  when tab1.program_type_id = applications.program_type_id then 490 when 
tab1.postal=    applications.postal then 1000 end )from tab1, applications where 
tab1.a = applications.a or tab1.b = applications.b

当然，您需要将此查询中的列名更改为您的列名。