Mysql 具有多个左联接的查询,未获得所需结果
这里有一个 有人能告诉我如何使用Mysql 具有多个左联接的查询,未获得所需结果,mysql,sql,left-join,Mysql,Sql,Left Join,这里有一个 有人能告诉我如何使用左连接获得此输出吗 notification_recipient pm_sender msg modification_page_id Peter Tom Hello NULL notification_recipient pm_sender msg modification_page_id Peter NULL
左连接
获得此输出吗
notification_recipient pm_sender msg modification_page_id
Peter Tom Hello NULL
notification_recipient pm_sender msg modification_page_id
Peter NULL NULL 2
以下是我尝试过的查询:
SELECT u.name AS notification_recipient,us.name AS pm_sender,
p.msg,um.page_id AS modification_page_id
FROM notification n
LEFT JOIN pm p ON p.pm_id = n.pm_id
LEFT JOIN users u ON u.user_id = p.recipent_id
LEFT JOIN users us ON us.user_id = p.sender_id
LEFT JOIN user_modification um ON um.modification_id = n.modification_id
WHERE u.name = 'Peter'
AND n.is_read = '0'
我在寻找某种排序条件联接,这意味着根据字段中是否存在值来联接不同的表,但在我的示例中找不到适用的联接。任何其他有效的解决方案也将不胜感激
背景:
我计划制作一个通知系统,向用户发送不同类型的消息(用户之间的私人消息、他们对条目的修改已被批准的消息等)。
当一个用户登录时,我想做一个查询,看看是否有该用户的未读通知。如果是,将通过Ajax向他发送通知
举例来说,假设Tom向Peter发送了一条私人消息,并且他对条目的修改得到批准,则将调用表pm
和user\u modification
中的两个触发器,将两个新行添加到通知中。列pm\u id
由pm
引用,modification\u id
由modification
引用<代码>已读取
默认为未读取
以下是表架构:
CREATE TABLE notification
(`id` int, `modification_id` int,`pm_id` int,`is_read` int)
;
INSERT INTO notification
(`id`,`modification_id`,`pm_id`,`is_read`)
VALUES
(1,1,NULL,0),
(2,NULL,1,0),
(3,2,NULL,0)
;
CREATE TABLE user_modification
(`modification_id` int, `user_id` int,`page_id` int, `is_approved` int)
;
INSERT INTO user_modification
(`modification_id`,`user_id`,`page_id`,`is_approved`)
VALUES
(1,1,5,1),
(2,2,2,1),
(3,3,3,0)
;
CREATE TABLE pm
(`pm_id` int, `sender_id` int,`recipent_id` int,`msg` varchar(200))
;
INSERT INTO pm
(`pm_id`,`sender_id`,`recipent_id`,`msg`)
VALUES
(1,1,2,'Hello');
CREATE TABLE users
(`user_id` int, `name`varchar(20))
;
INSERT INTO users
(`user_id`,`name`)
VALUES
(1,'Tom'),
(2,'Peter'),
(3,'David')
;
如果用户David登录,我希望得到一个通知的输出。每种消息类型的每一行
notification_recipient pm_sender msg modification_page_id
Peter Tom Hello NULL
notification_recipient pm_sender msg modification_page_id
Peter NULL NULL 2
给Peter的通知如下:
`1.You have received a private message from Tom.
2.your modification on <a href='mysite.com/5'>that page</a> is approved`.
`1.您收到了汤姆的私人信息。
2.您对“”的修改已批准。
此查询应该可以完成此任务。这是
SELECT
n.id,
IF(pmu.name IS NULL, pmm.name, pmu.name) recipient,
pmus.name sender, pm.msg, m.modification_id
FROM
notification n
LEFT JOIN user_modification m ON (n.modification_id = m.modification_id)
LEFT JOIN pm ON (n.pm_id = pm.pm_id)
LEFT JOIN users pmu ON (pm.recipent_id = pmu.user_id)
LEFT JOIN users pmus ON (pm.sender_id = pmus.user_id)
LEFT JOIN users pmm ON (m.user_id = pmm.user_id)
WHERE
(pmu.name = 'Peter' OR
pmm.name = 'Peter') AND
n.is_read = 0;