Question

我有以下数据：

select * from art_skills_table;
+----+------+---------------------------+
| ID | Name | skills                    | 
+----+------+---------------------------|
|  1 | Anna | ["painting","photography"]| 
|  2 | Bob  | ["drawing","sculpting"]   |
|  3 | Cat  | ["pastel"]                |
+----+------+---------------------------+

select * from computer_table;
+------+------+-------------------------+
| ID   | Name | skills                  |  
+------+------+-------------------------+
|    1 | Anna |   ["word","typing"]     |
|    2 | Cat  |   ["code","editing"]    |
|    3 | Bob  |   ["excel","code"]      |
+------+------+-------------------------+

我想编写一条产生下表的SQL语句。

+------+------+-----------------------------------------------+
| ID   | Name | skills                                        |  
+------+------+-----------------------------------------------+
|    1 | Anna |   ["painting","photography","word","typing"]  |
|    2 | Bob  |   ["drawing","sculpting","excel","code"]      |
|    3 | Cat  |   ["pastel","code","editing"]                 |
+------+------+-----------------------------------------------+

我尝试过类似SELECT * from art_skills_table LEFT JOIN computer_table ON name的操作。但是，它不能满足我的需求。我已经读过array_cat了，但是在实现它时遇到了一些麻烦。

Answer 1

首先，我想知道为什么将数据存储在这样的模型中。认为NoSQL数据库缺乏联接和...的能力。

...语义三元组将采用主语-谓语-宾语的形式。
...一个键值（KV）存储使用关联数组。
...一个关系数据库将被规范化。

有关用例的一些信息会有所帮助。

尽管如此，您可以使用CONCAT和REPLACE选择所需格式的数据。

SELECT art_skills_table.ID, computer_table.name, 
CONCAT(
  REPLACE(art_skills_table.skills, '}',','), 
  REPLACE(computer_table.skills, '{','')
)
FROM art_skills_table JOIN computer_table ON art_skills_table.ID = computer_table.ID

查询返回以下结果：

+----+------+--------------------------------------------+
| ID | Name |                   Skills                   |
+----+------+--------------------------------------------+
|  1 | Anna | {"painting","photography","word","typing"} |
|  2 | Cat  | {"drawing","sculpting","code","editing"}   |
|  3 | Bob  | {"pastel","excel","code"}                  |
+----+------+--------------------------------------------+

即使Bob的值不同，我也已将ID用于JOIN。 JOIN可能应该在名称上完成。

JOIN computer_table ON art_skills_table.Name = computer_table.Name

顺便说一句，您需要告诉我们您正在运行哪种SQL引擎。

如何选择带有串联列的表？

1 个答案: