我是关系型数据库的新手...... 我想列出同一个DB中许多表中的所有数据。
我有一个包含其他表的NAMES的表。像这样:
mysql> select * from CarKeys;
+-------+
| Name |
+-------+
| Audi |
| Honda |
+-------+
mysql> select * from Audi;
+-------+---------+--------+
| Model | NumDoor | Colour |
+-------+---------+--------+
| A4 | 4 | White |
+-------+---------+--------+
mysql> select * from Honda;
+------------+---------+--------+
| Model | NumDoor | Colour |
+------------+---------+--------+
| Civic | 4 | White |
| Civic | 4 | White |
| HomdaModel | 5 | Red |
+------------+---------+--------+
但现在,我不想分别致电select * from Honda
和select * from Audi
。
我想要一个select * from
CarKeys
表
SELECT * FROM (SELECT * FROM CarKeys);
命令
我试过这个:
ERROR 1248 (42000): Every derived table must have its own alias
但是我收到了一条错误消息:
WebApp
如何获取此数据库中所有表的所有信息?我使用的架构的名称是{{1}}
答案 0 :(得分:1)
您的SQL设计存在很大问题。你不应该这样叫表。为什么不为所有汽车创建一个表并添加一个列,引用包含名称的表?
+-------+-------+
| ID | Name |
+-------+-------+
| 1 | Audi |
| 2 | Honda |
+-------+-------+
+------------+---------+--------+--------+
| Model | NumDoor | Colour | Brand |
+------------+---------+--------+--------+
| Civic | 4 | White | 2 |
| A4 | 4 | White | 1 |
| Civic | 4 | White | 2 |
| HomdaModel | 5 | Red | 2 |
+------------+---------+--------+--------+
答案 1 :(得分:1)
这个表没有很好的关系,你可以尝试:
SELECT * FROM Audi
UNION
SELECT * FROM Honda
结果:
+------------+---------+--------+
| Model | NumDoor | Colour |
+------------+---------+--------+
| A4 | 4 | White |
| Civic | 4 | White |
| Civic | 4 | White |
| HomdaModel | 5 | Red |
+------------+---------+--------+
答案 2 :(得分:0)