如何按用户ID加入3个表并返回存在链接的记录？

Question

我有3张桌子：

用户

• 用户ID
• 用户名

约会

• Appointid
• useridfk
• Appointname
• 开始
• 结束

plinks

• PID
• useridfk
• 链接
• projectidfk

我想要做的是通过userid加入所有表并显示plink中存在链接的所有约会，其中users username = Mike

Answer 1

尝试这样的事情：

select b.* from users a join Appointments b on a.userid=b.useridfk join plinks c on a.userid=c.useridfk where a.username='Mike' and c.plinks is not null

Answer 2

你尝试过这样的事吗？

SELECT * 
FROM Appointments
INNER JOIN users
ON users.userid = Appointments.useridfk
INNER JOIN plinks
ON users.userid = plinks.useridfk
WHERE users.username = 'Mike'
AND plinks.link IS NOT NULL;

Answer 3

猜测表之间的关系我会这样做：

//In below code, you can replace input[type=text] with classname/id of input field. 
$('input[type=text]').on('change', function() {  
    alert($('input[type=text]').val());
});

Answer 4

试试这个

select a.* from Appointments a inner join Users b on b.UserId = a.UserIdFk

内部联接PLinks c on b.UserId = c.UserIdFk 其中b.UserName ='Mike'且c.Link不为空

Answer 5

你可以尝试这解决了你的问题...

SELECT * FROM Appointments as a INNER JOIN users as u ON u.userid = a.useridfk INNER JOIN plinks as p ON u.userid = p.useridfk
WHERE u.username = 'Mike' AND p.link IS NOT NULL;