Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
690 views
in Technique[技术] by (71.8m points)

mysql - Joining different tables based on column value

I have a table called notifications:

CREATE TABLE `notifications` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `user_id` int(11) DEFAULT NULL,
  `type` varchar(20) NOT NULL DEFAULT '',
  `parent_id` int(11) DEFAULT NULL,
  `parent_type` varchar(15) DEFAULT NULL,
  `type_id` int(11) DEFAULT NULL,
  `etc` NULL
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=25 DEFAULT CHARSET=utf8;

Each notification is related to a different table, the value of parent_type field specifies the name of the table that I want to * join the table with. All target tables have several similar columns:

CREATE TABLE `tablename` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `is_visible` tinyint(1) NOT NULL,      
  `etc` NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8;

Currently I'm using this query for selecting notifcations that their related row in the target table exists and their is_visible field is 1:

SELECT n.id, 
FROM notifications n 
LEFT JOIN books b ON n.parent_id = b.id AND n.parent_type = 'book' AND b.is_visible = 1
LEFT JOIN interviews i ON n.parent_id = i.id AND n.parent_type = 'interview' AND i.is_visible = 1
LEFT JOIN other tables...
WHERE n.user_id = 1
GROUP BY n.id

But since it is a LEFT JOIN it returns the notification if it matches any table or not, how can I rewrite it so it doesn't return notifications that don't match with any row in the target table? I have also tried the CASE statement unsuccessfully.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I'm not 100% sure the syntax is right and I have no chance to test it right now, but the idea should be clear.

SELECT DISTINCT n.id 
FROM notifications n 
JOIN (
     (SELECT b.id, 'book' AS type FROM books b WHERE b.is_visible = 1)
  UNION
     (SELECT i.id, 'interview' AS type FROM interviews i WHERE i.is_visible = 1)
) ids ON n.parent_id = ids.id AND n.parent_type = ids.type
WHERE n.user_id = 1

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...