Question

SQL需要帮助的初学者就是问题所在。对于我在postgresql上测试的任何数据库，可以是SQL stmt。我有2个SQL表的以下问题：

表编号1

 .....................  
|Boy Name| Girl Name |  
|--------------------|  
|Michael |  Anne     |  
|Michael |  Misty    |  
|Michael |  Simone   |  
|Michael |  Diane    |  
|Michael |  Ariel    |  
|Jack    |  Misty    |  
|Jack    |  Simone   |  
|Jack    |  Anne     |  
|Sam     |  Simone   |  
|Sam     |  Misty    |  
|Sam     |  Anne     |  
|Sam     |  Mini     |  
|Sam     |  Valery   |   
----------------------

表编号2

..................................
|Boy Name | Anne | Misty | Simone|
|--------------------------------|
|Michael  |  yes | no    |  no   | 
|Jack     |  yes | yes   |  yes  | 
|Sam      |  no  | no    |  yes  | 
..................................

（IDK就像一个问题：一个女孩喜欢特定的男孩。可以是，不，或没有。）

期望的结果

查询返回2列的表： - 男孩通过2号桌从女孩身上获得的每一个“不”的行。

.....................
|Boy Name| Girl Name |
|--------------------|
|Michael |  Misty    |
|Michael |  Simone   |
|Sam     |  Misty    |
|Sam     |  Anne     |
----------------------

您可以将结果视为一个新行a，其中包含男孩名称的列，另一个名称为女孩的名称，表格编号为2中为“否”。

Answer 1

首先，我建议使用不同的表格设计：

CREATE TABLE person
(
 id INT PRIMARY KEY,
 person_name VARCHAR(200),
 sex CHAR(1)
);
-- possible record: 1, 'John', 'M'; 2, 'Mary', 'F';

CREATE TABLE person_symphaty
(
 person_id INT, -- FK to person table
 symphaty_person_id INT -- FK to person table
);
-- you will store here if the person likes another one. 1 point here is that you can store that person doesn't like somebody, then you need some flag, for example is_symphaty INT (then 1 will mean likes, 0 will mean dislike)

然后查询本身

    SELECT pm.name AS male_name, pf.name AS female_name
    FROM person pm
    CROSS JOIN person pf
    WHERE pm.sex = 'M' AND pf.sex = 'F'
NOT EXISTS (
    SELECT * FROM person_symphaty ps WHERE ps.person_id = pm.id AND ps.symphaty_person_id = pf.id
    )

可能会出现错误，因为我根本没有测试这些查询。

SQL查询由另一个表条件限制

1 个答案: