Mysql SQL从3个表中选择并合并列名
我试图从MySQL中的3个表中获取数据,并更改/合并它们的列名。现在,当我使用Mysql SQL从3个表中选择并合并列名,mysql,sql,Mysql,Sql,我试图从MySQL中的3个表中获取数据,并更改/合并它们的列名。现在,当我使用将列名设置为时,它们是重复的 人员表格: id applicant_id employee_id --------------------------------- 1 3 6 2 4 10 3 12 30 id applicant_
将列名设置为
时,它们是重复的
人员表格:
id applicant_id employee_id
---------------------------------
1 3 6
2 4 10
3 12 30
id applicant_id applicant_note applicant_note_date
-----------------------------------------------------------
1 3 "Was good" 2013-05-01
1 4 "Was so-so" 2013-06-07
2 4 "Was bad" 2013-06-08
3 4 "Was great" 2013-06-10
id employee_id employee_note employee_note_date
--------------------------------------------------------
1 10 "Was ok" 2013-07-20
1 10 "Was great" 2013-07-21
2 30 "Was bad" 2013-08-01
3 30 "Was so-so" 2013-08-02
申请人表格:
id applicant_id employee_id
---------------------------------
1 3 6
2 4 10
3 12 30
id applicant_id applicant_note applicant_note_date
-----------------------------------------------------------
1 3 "Was good" 2013-05-01
1 4 "Was so-so" 2013-06-07
2 4 "Was bad" 2013-06-08
3 4 "Was great" 2013-06-10
id employee_id employee_note employee_note_date
--------------------------------------------------------
1 10 "Was ok" 2013-07-20
1 10 "Was great" 2013-07-21
2 30 "Was bad" 2013-08-01
3 30 "Was so-so" 2013-08-02
员工表格:
id applicant_id employee_id
---------------------------------
1 3 6
2 4 10
3 12 30
id applicant_id applicant_note applicant_note_date
-----------------------------------------------------------
1 3 "Was good" 2013-05-01
1 4 "Was so-so" 2013-06-07
2 4 "Was bad" 2013-06-08
3 4 "Was great" 2013-06-10
id employee_id employee_note employee_note_date
--------------------------------------------------------
1 10 "Was ok" 2013-07-20
1 10 "Was great" 2013-07-21
2 30 "Was bad" 2013-08-01
3 30 "Was so-so" 2013-08-02
我只有员工id
。我想确保我从员工和申请人那里得到了所有的笔记,我想让他们合并到同一列中,而不是有重复的NULL
值的列。我想返回如下结果:
note date type
------------------------------------------------
"Was so-so" 2013-06-07 applicant
"Was bad" 2013-06-08 applicant
"Was great" 2013-06-10 applicant
"Was ok" 2013-07-20 employee
"Was great" 2013-07-21 employee
我现在的处境是:
SELECT
applicants.applicant_note AS note,
applicants.applicant_note_date AS date,
employees.employee_note AS note,
employees.employee_note_date AS date
IF(applicants.applicant_id IS NULL, 'employee', 'applicant') as type
FROM
employees
JOIN
people
ON
people.employee_id = employees.employee_id
JOIN
applicants
ON
applicants.applicant_id = people.applicant_id
WHERE
employees.employee_id = 10
有没有一种方法可以仅使用SQL来实现这一点?或者我需要运行单独的查询来获取申请人id和员工id吗?您需要使用
UNION ALL
SELECT employee_note note,
employee_note_date date,
'employee' type
FROM people a
INNER JOIN employees b
ON a.employee_ID = b.employee_ID
WHERE a.employee_ID = 10
UNION ALL
SELECT applicant_note note,
applicant_note_date date,
'applicant' type
FROM people a
INNER JOIN applicants b
ON a.applicant_id = b.applicant_id
WHERE a.employee_ID = 10
SELECT employee_note note,
employee_note_date date,
'employee' type
FROM employees e
WHERE e.employee_id = 10
UNION
SELECT applicant_note note,
applicant_note_date date,
'applicant' type
FROM applicants a
WHERE applicant_id = (SELECT applicant_id FROM people WHERE employee_id = 10)
使用JOIN的一个优点是,查询可以转换为一个派生表,允许在查询末尾的单个位置指示员工id(或者通过申请人id或people.id限制列表,只需稍加修改)。此外,该表还可以包含people.id,以确保在查询中显示正确的人。例如:
SELECT * FROM (
SELECT p.id person,
employee_note note,
employee_note_date date,
'employee' type
FROM employees e
JOIN people p on p.employee_id = e.employee_id
UNION
SELECT p.id person,
applicant_note note,
applicant_note_date date,
'applicant' type
FROM applicants a
JOIN people p on p.applicant_id = a.applicant_id
) q
WHERE q.person in(SELECT id FROM people WHERE employee_id = 10)
下面是创建这三个表的语句
CREATE TABLE `people` (`id` int(11) NOT NULL, `applicant_id` int(11) NOT NULL,
`employee_id` int(11) NOT NULL, PRIMARY KEY (`id`),
KEY `applicant_id_index` (`applicant_id`),
KEY `employee_id_index` (`employee_id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE `employees` (`employees_id` int(11) NOT NULL AUTO_INCREMENT,
`id` int(11) NOT NULL, `employee_id` int(11) NOT NULL,
`employee_note` varchar(12) DEFAULT NULL,
`employee_note_date` datetime DEFAULT NULL,
PRIMARY KEY (`employees_id`),
KEY `employee_id_index` (`employee_id`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
CREATE TABLE `applicants` (
`applicants_id` int(11) NOT NULL AUTO_INCREMENT,
`id` int(11) NOT NULL,
`applicant_id` int(11) NOT NULL,
`applicant_note` varchar(12) DEFAULT NULL,
`applicant_note_date` datetime DEFAULT NULL,
PRIMARY KEY (`applicants_id`),
KEY `applicant_id_index` (`applicant_id`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
INSERT INTO `employees` VALUES (1,1,10,'Was ok','2013-07-20 00:00:00'),
(2,1,10,'Was great','2013-07-21 00:00:00'),(3,2,30,'Was bad','2013-08-01 00:00:00'),
(4,3,30,'Was so-so','2013-08-02 00:00:00');
INSERT INTO `people` VALUES (1,3,6),(2,4,10),(3,12,30);
INSERT INTO `applicants` VALUES (1,1,3,'Was good','2013-05-01 00:00:00'),
(2,1,4,'Was so-so','2013-06-07 00:00:00'),
(3,2,4,'Was bad','2013-06-08 00:00:00'),
(4,3,4,'Was great','2013-06-10 00:00:00');
杰出的但是,在第一次选择时,您是否可以跳过
内部联接
,根据notes表中的员工id执行WHERE
?我试过了,它似乎工作得很好,只是好奇地听到你使用内部连接的理由