Question

我相信我在这里要实现的是嵌套查询 - 但我不确定如何继续。

我有2个表用户和通知我正在查询通知表以获取详细信息，但我也是 LEFT OUTER JOIN 以获取我的用户信息。我拥有的是这样的：

          USERS TABLE     
| user id | username | user image |


    NOTIFICATIONS TABLE  
| receiver id | sender id |

获取接收器信息不是问题，因为我使用以下查询：

SELECT
    notifications.senderID, notifications.receiverID, users.username, users.image
FROM
    notifications
    LEFT OUTER JOIN users ON users.id = notifications.receiverID
WHERE notifications.receiverID = ' xxx '

我需要添加的内容是一种foreach记录的方法，也可以获取 THAT 记录的users.username和users.image（将对应于users.id为notifications.senderID ）以及主记录（ notifications.recieverID ）

如果可能我想在一个查询中以及PHP foreach循环之外保留它。

Answer 1

使用不同的别名

加入users表两次

SELECT notifications.senderID, notifications.receiverID, 
       receiver.username as receiver_name, receiver.image as receiver_image,
       sender.username as sender_name, sender.image as sender_image
FROM notifications
LEFT OUTER JOIN users as receiver ON receiver.id = notifications.receiverID
LEFT OUTER JOIN users as sender ON sender.id = notifications.senderID
WHERE notifications.receiverID = ' xxx '

Answer 2

这是查询，例如

选择 User。id，User。username，User。password，User。email，{{ 1}}。User，role。User，created。User，modified。Userdetail，{{1} }。user_id，Userdetail，first_name。last name，Userdetail。address 来自Userdetail。telephone AS cakephp_test LEFT JOIN users。User AS cakephp_test ON（userdetails。Userdetail = Userdetail。user_id）WHERE User。id = 51 LIMIT 1

在你的情况下

SELECT notifications.senderID，notifications.receiverID，users.username，users.image FROM database_name.users AS User LEFT JOIN database_name.notification AS notification ON（users.id = notifications.receiverID）WHERE User.id ='xxx'限制1

MySQL选择所有记录并为每个选择单独的记录 - 在一个查询中

2 个答案: