Question

我不是程序员，我从这个表格中读到了很多关于如何解决我的问题，但我的搜索并不好

我有两张桌子

表1：成员

id*| name | surname
-------------------
1  | Joe  | Smith
2  | Mary | Sullivan
3  | Will | Stevenson

表2：消息

---------------------------------
id_message*| from | to | message
---------------------------------
       1   |   2  |  1 | test
       2   |   1  |  2 | re:test
       3   |   3  |  1 | hi

*自动增加字段

我希望进行查询，列出所有消息，如下所示：

Mary Sullivan  | Joe Smith     | test 
Joe Smith      | Mary Sullivan | re:test
Will Stevenson | Joe Smith     | hi

我真的真的迷路了有人可以帮忙吗？谢谢！

Answer 1

您需要使用members

加入messages表2次

select
concat(mem1.name,' ',mem1.surname) as `from_name`,
concat(mem2.name,' ',mem2.surname) as `to_name`,
m.message 
from messages m
join members mem1 on mem1.id = m.`from`
join members mem2 on mem2.id = m.`to`

Answer 2

尝试：

select from1.name+" "+from1.surname, to2.name+" "+to2.surname, message from table2 
join table1 from1 on table2.`from` = table1.id
join table1 to2 on table2.`to` = table1.id

Answer 3

你需要写一个Select with alias来引用成员表2次：

SELECT CONCAT(M1.surname, ' ', M1.name) as FROM, CONCAT( M2.surname, ' ', M2.name) as TO FROM 
members M1 INNER JOIN 
messages M on M.from = M1.id 
INNER JOIN messages M2 on M.to = M2.id

表1具有名称和姓氏，表2具有两列引用表1中的名称

3 个答案: