TABLE PEDIDO
id_name | ID_cabimento | ID_direction
1 | 4 | 5
2 | 3 | 6
3 | 4 | 5
TABLE USER
id_name | name
1 | João
2 | Maria
3 | António
4 | Manuel
我想要结果
name | cabimento | direction
João | Manuel | Tozé
Maria | António | Joaquim
António | Manuel | Tozé
...
我尝试了UNION和JOIN,但没有得到理想的结果...因为我只能解码1个ID。
答案 0 :(得分:1)
你可以在同一张桌子上加入n次,只需使用n个别名
select n.name as name, c.name as cabimento, d.name as direction
from pedido p
inner join user n on p.id_name = n.id_name
inner join user c on p.id_name = c.id_cabimento
inner join user d on p.id_name = d.id_direction