Question

Hello People我有两张桌子......

Users
- id (
- username

sample data (1:alex,2:rocky,3:kothi)



linked_table
- from_id
- to_id
sample data (1:2,1:3)

我想从linked_table.i中获取并显示from_name和To_name尝试加入并选择，但它只从from_name

获取

    $Get_Linked_name = \DB::table('linked_table')
    ->join('users','linked_table.from_id','=','users.Id')
    ->where(array('linked_table.from_id'=>1))
    ->select()->get();

'linked_table.from_id'=>1已链接到2,3现在我想要获取username 2,3我知道可以使用另一个空数组和foreach loop完成但是还有其他更好的选择吗？

updated: From_name and To_name are nothing but username of users table

Answer 1

使用UNION

SELECT u.id, u.username
FROM Users u, JOIN linked_table lt
ON u.id = lt.from_id
WHERE lt.from_id = 1
UNION
SELECT u.id, u.username
FROM Users u, JOIN linked_table lt
ON u.id = lt.to_id
WHERE lt.from_id = 1

Answer 2

您可以尝试此查询：

SELECT from.username as From_name, to.username as To_name
FROM   Users from, Users to, linked_table lt
WHERE  from.id = lt.from_id
       AND to.id = lt.to_id
       AND from.id = 1

需要Sql逻辑

2 个答案: