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部門表からすべての部門詳細を選択したい。しかし、各 head_user_id、assistant_1_user_id および assistant_2_user_id の name_full が必要です

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);
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2 に答える 2

1

これを試して、

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`

お役に立てれば。

ちなみに、制約は次のようになります

  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)
于 2012-09-10T15:50:55.550 に答える
1 
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
于 2012-09-10T16:00:13.440 に答える