Question

我想从分区表中选择所有分区详细信息。但我想要每个head_user_id的name_full，assistant_1_user_id和assistant_2_user_id

CREATE TABLE IF NOT EXISTS `user_mst` (
  `user_id` int(11) NOT NULL AUTO_INCREMENT,
  `name_full` varchar(250) DEFAULT NULL,
  `name_with_initials` varchar(250) DEFAULT NULL,
  `gender` char(1) NOT NULL,
  `nic_no` varchar(20) DEFAULT NULL,
  `title_id` varchar(10) NOT NULL,
  `dob` date DEFAULT NULL,
  `address1` varchar(100) DEFAULT NULL,
  `address2` varchar(100) DEFAULT NULL,
  `address3` varchar(100) DEFAULT NULL,
  `address4` varchar(100) DEFAULT NULL,
  `tele_no` varchar(40) DEFAULT NULL,
  `mob_no` varchar(40) DEFAULT NULL,
  `email_personal` varchar(100) DEFAULT NULL,
  `role_id` varchar(10) NOT NULL,
  `creator_user_id` varchar(50) NOT NULL,
  `active_user` tinyint(1) NOT NULL,
  `emp_no1` varchar(50) DEFAULT NULL,
  `emp_no2` varchar(50) DEFAULT NULL,
  `unit_id` varchar(50) DEFAULT NULL,
  `division_id` varchar(50) DEFAULT NULL,
  `username` varchar(50) NOT NULL,
  `password` varchar(50) NOT NULL,
  `email_official` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`user_id`),
  KEY `fk_user_mst_title_mst` (`title_id`),
  KEY `fk_user_mst_role_mst1` (`role_id`),
  KEY `fk_user_mst_unit_mst1` (`unit_id`,`division_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ;

INSERT INTO `user_mst` (`user_id`, `name_full`, `name_with_initials`, `gender`, `nic_no`, `title_id`, `dob`, `address1`, `address2`, `address3`, `address4`, `tele_no`, `mob_no`, `email_personal`, `role_id`, `creator_user_id`, `active_user`, `emp_no1`, `emp_no2`, `unit_id`, `division_id`, `username`, `password`, `email_official`) VALUES
(2, 'dss', 'ddsf', 'm', '2353246565v', '12', '2012-03-12', 'ewtewt', 'ryery', 'ertert', 'wetwet', '325235325', '23523534523', 'fdrg@yahoo.com', '1', '1', 1, '', NULL, '1', '1', 'cde', '123', 'sasas@gmail.com'),
(3, 'wrwer', 'egrt', 'f', '2432544663', '12', '2012-03-26', 'erwerw', 'wetw', 'ewtwe', 'ewtw', '132435435', '1243345345', 'dfggdfg', '23', '1', 1, '12', '12', '12', '3', 'pas', '123', 'sdasda'),
(4, 'asd', 'asd', 'f', '5671234676V', '12', '2012-03-05', 'sdgdsgsd', 'sdgsdgds', 'rgwergwetg', 'ergry', '12141242145', '1242135346', 'prameeshas@yahoo.com', '1', '123567', 1, '1234', '123', '1', '1', 'abc', '234', 'prameeshas@yahoo.com');    

  CREATE TABLE IF NOT EXISTS `division_mst` (
  `division_id` varchar(50) NOT NULL,
  `division_code` varchar(50) NOT NULL,
  `name` varchar(100) NOT NULL,
  `description` varchar(500) DEFAULT NULL,
  `colour` varchar(50) DEFAULT NULL,
  `head_user_id` int(11) DEFAULT NULL,
  `assistant_1_user_id` int(11) DEFAULT NULL,
  `assistant_2_user_id` int(11) DEFAULT NULL,
  `main_division_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`division_id`),
  KEY `fk_division_mst_user_mst1` (`head_user_id`),
  KEY `fk_division_mst_user_mst2` (`assistant_1_user_id`),
  KEY `fk_division_mst_user_mst3` (`assistant_2_user_id`),
  KEY `fk_division_mst_main_division_mst1` (`main_division_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `division_mst` (`division_id`, `division_code`, `name`, `description`, `colour`, `head_user_id`, `assistant_1_user_id`, `assistant_2_user_id`, `main_division_id`) VALUES
('1', 'D001', 'Administration', 'tjrtujrt', 'pink', 1, 2, NULL, 1),
('2', 'D002\n', 'Human Resource\n', 'yjghkhk', 'red', 1, 3, 2, 1),
('3', 'D003', 'Marketing', 'jhghfg', 'green', 2, 1, 3, 2),
('4', 'D004', 'IT ', NULL, NULL, NULL, NULL, NULL, NULL),
('5', 'D005', 'Accounting ', NULL, NULL, NULL, NULL, NULL, NULL);

Answer 1

试试这个，

SELECT  b.`name_full` head_user,
        c.`name_full` assistant_1,
        d.`name_full assistant_2
FROM    `division_mst` a
        INNER JOIN `user_mst` b
            ON a.`head_user_id` = b.`user_id`
        INNER JOIN `user_mst` c
            ON a.`assistant_1_user_id` = b.`user_id`
        INNER JOIN `user_mst` d
            ON a.`assistant_2_user_id` = b.`user_id`

希望这有帮助。

顺便说一下，你的约束应该是这样的

  CONSTRAINT `fk_division_mst_user_mst1` FOREIGN KEY (`head_user_id`) REFERENCES `user_mst`(user_id),
  CONSTRAINT `fk_division_mst_user_mst2` FOREIGN KEY (`assistant_1_user_id`) REFERENCES `user_mst`(user_id),
  CONSTRAINT `fk_division_mst_user_mst3` FOREIGN KEY (`assistant_2_user_id`) REFERENCES `user_mst`(user_id)

Answer 2

select 
    d.*, 
    u1.name_full AS head_user, 
    u2.name_full AS assistant1, 
    u3.name_full AS assistant2
from division_mst d 
    LEFT OUTER JOIN user_mst u1 ON d.head_user_id=u1.user_id
    LEFT OUTER JOIN user_mst u2 ON d.assistant_1_user_id=u2.user_id
    LEFT OUTER JOIN user_mst u3 ON d.assistant_2_user_id=u3.user_id

将具有三个外键的MySQL表连接到同一个表

2 个答案: