【发布时间】:2015-02-25 13:37:28
【问题描述】:
我有一张联系表。每个联系人可以有多个电话号码,具有不同或相同的表。同样,每个联系人可以拥有多封具有不同或相同标签的电子邮件。我需要选择属于特定联系人的所有不同电话号码和电子邮件 ID。
CREATE TABLE IF NOT EXISTS `contact` (
`id` int(8) unsigned NOT NULL ,
`fname` varchar(64) NOT NULL,
`lname` varchar(64) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 ;
CREATE TABLE IF NOT EXISTS `phone` (
`id` int(8) unsigned NOT NULL ,
`sourceid` int(8) unsigned NOT NULL ,
`type` varchar(16) NOT NULL,
`phone` varchar(16) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 ;
CREATE TABLE IF NOT EXISTS `email` (
`id` int(8) unsigned NOT NULL ,
`sourceid` int(8) unsigned NOT NULL ,
`type` varchar(16) NOT NULL,
`email` varchar(128) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 ;
INSERT INTO contact values ( 1,'john' ,'j' ),
(2, 'jose' ,'f' ),
(3, 'test' ,'k' ),
(4, 'tester' ,'j' );
INSERT INTO phone values( 1 ,1, 'Home', '123456' ),
( 2 ,1, 'Home', '123456342' ),
( 3 ,1, 'Office', '12345645' ),
( 4 ,1, 'Mobile', '1234567' ),
( 5 ,2, 'Home', '123456' ),
( 5 ,2, 'Home', '987556' );
INSERT INTO email values
( 1 ,1, 'Home', 'john@gmail.com' ),
( 2 ,1, 'Home', 'john@yahoo.com' ),
( 3 ,1, 'Office', 'john@inc.com' ),
( 4 ,2, 'Home', 'jose@gmail.com' ),
( 5 ,4, 'Home', 'test@test.com' );
我尝试了 GROUP_CONCAT,但它的标签条目重复。
SELECT C.id, fname, lname
, GROUP_CONCAT(ph.sourceid),
GROUP_CONCAT(em.sourceid),
GROUP_CONCAT(ph.type), GROUP_CONCAT(ph.phone),
GROUP_CONCAT(em.type), GROUP_CONCAT(em.email)
FROM contact AS C
LEFT JOIN phone AS PH ON PH.sourceid = C.id
LEFT JOIN email EM ON EM.sourceid = C.id
WHERE C.id='1'
GROUP BY C.id
【问题讨论】:
标签: mysql sql select left-join group-concat