当玛格丽特担任经理时，我需要获取乔治·费斯托的薪水数据。
这是位置（标题表）表：
___________________________________________________
| name | title | from_date | to_date |
----------------------------------------------------
| Margareta | Manager | 1985-01-01 | 1991-10-01 |
这是员工薪水表：
___________________________________________________
| name | salary | from_date | to_date |
----------------------------------------------------
Georgi Facello 60117 1986-06-26 1987-06-26
Georgi Facello 62102 1987-06-26 1988-06-25
Georgi Facello 66074 1988-06-25 1989-06-25
Georgi Facello 66596 1989-06-25 1990-06-25
Georgi Facello 66961 1990-06-25 1991-06-25
Georgi Facello 71046 1991-06-25 1992-06-24
Georgi Facello 74333 1992-06-24 1993-06-24
Georgi Facello 75286 1993-06-24 1994-06-24
Georgi Facello 75994 1994-06-24 1995-06-24
Georgi Facello 76884 1995-06-24 1996-06-23
Georgi Facello 80013 1996-06-23 1997-06-23
Georgi Facello 81025 1997-06-23 1998-06-23
Georgi Facello 81097 1998-06-23 1999-06-23
Georgi Facello 84917 1999-06-23 2000-06-22
Georgi Facello 85112 2000-06-22 2001-06-22
Georgi Facello 85097 2001-06-22 2002-06-22
Georgi Facello 88958 2002-06-22 9999-01-01
所以我的查询是
SELECT employees.name, salaries.*
FROM salaries
INNER JOIN employees ON employees.emp_no = salaries.emp_no
WHERE salaries.from_date >= (
SELECT titles.from_date
FROM titles
JOIN employees ON employees.emp_no = titles.emp_no
WHERE titles.title = 'manager' AND employees.name = 'margareta markovitch'
)
AND salaries.to_date <= (
SELECT titles.to_date
FROM titles
JOIN employees ON employees.emp_no = titles.emp_no
WHERE titles.title = 'manager' AND employees.name = 'margareta markovitch'
)
这是结果
___________________________________________________
| name | salary | from_date | to_date |
----------------------------------------------------
Georgi Facello 60117 1986-06-26 1987-06-26
Georgi Facello 62102 1987-06-26 1988-06-25
Georgi Facello 66074 1988-06-25 1989-06-25
Georgi Facello 66596 1989-06-25 1990-06-25
Georgi Facello 66961 1990-06-25 1991-06-25
问题是结果中不包含一行：
Georgi Facello 71046 1991-06-25 1992-06-24
哪一个仍然有效，因为玛格丽塔（Margareta）直到91年八月仍是经理
您对这个问题有建议吗？
答案 0 :(得分：1)
您非常亲密。只需要包括所有重叠的时期：
SELECT DISTINCT employees.name, salaries.*
FROM salaries
INNER JOIN employees ON employees.emp_no = salaries.emp_no
WHERE salaries.from_date < (
SELECT titles.to_date
FROM titles
JOIN employees ON employees.emp_no = titles.emp_no
WHERE titles.title = 'manager' AND employees.name = 'margareta markovitch'
)
AND salaries.to_date > (
SELECT titles.from_date
FROM titles
JOIN employees ON employees.emp_no = titles.emp_no
WHERE titles.title = 'manager' AND employees.name = 'margareta markovitch'
)
答案 1 :(得分：1)
对PeterHe的替代答案，如果您只担心玛格丽塔担任经理期间是否开始一年，则仅与from_date进行比较，而忽略to_date：
SELECT employees.name, salaries.*
FROM salaries
INNER JOIN employees ON employees.emp_no = salaries.emp_no
WHERE salaries.from_date >= (
SELECT titles.from_date
FROM titles
JOIN employees ON employees.emp_no = titles.emp_no
WHERE titles.title = 'manager' AND employees.name = 'margareta markovitch'
)
AND salaries.from_date <= (
SELECT titles.to_date
FROM titles
JOIN employees ON employees.emp_no = titles.emp_no
WHERE titles.title = 'manager' AND employees.name = 'margareta markovitch'
)
这将返回在玛格丽塔担任经理期间一年开始的所有条目。
答案 2 :(得分：1)
如果您需要的代码更少...
DROP TABLE IF EXISTS titles;
CREATE TABLE titles
(name VARCHAR(12) NOT NULL
,title VARCHAR(12) NOT NULL
,from_date DATE NOT NULL
,to_date DATE NOT NULL
);
INSERT INTO titles VALUES
('Margareta','Manager','1985-01-01','1991-10-01');
DROP TABLE IF EXISTS salaries;
CREATE TABLE salaries
(name VARCHAR(30) NOT NULL
,salary INT NOT NULL
,from_date DATE NOT NULL
,to_date DATE NOT NULL
,PRIMARY KEY(name,from_date)
);
INSERT INTO salaries VALUES
('Georgi Fecello',60117,'1986-06-26','1987-06-26'),
('Georgi Fecello',62102,'1987-06-26','1988-06-25'),
('Georgi Fecello',66074,'1988-06-25','1989-06-25'),
('Georgi Fecello',66596,'1989-06-25','1990-06-25'),
('Georgi Fecello',66961,'1990-06-25','1991-06-25'),
('Georgi Fecello',71046,'1991-06-25','1992-06-24'),
('Georgi Fecello',74333,'1992-06-24','1993-06-24'),
('Georgi Fecello',75286,'1993-06-24','1994-06-24'),
('Georgi Fecello',75994,'1994-06-24','1995-06-24'),
('Georgi Fecello',76884,'1995-06-24','1996-06-23'),
('Georgi Fecello',80013,'1996-06-23','1997-06-23'),
('Georgi Fecello',81025,'1997-06-23','1998-06-23'),
('Georgi Fecello',81097,'1998-06-23','1999-06-23'),
('Georgi Fecello',84917,'1999-06-23','2000-06-22'),
('Georgi Fecello',85112,'2000-06-22','2001-06-22'),
('Georgi Fecello',85097,'2001-06-22','2002-06-22');
SELECT s.*
FROM titles t
JOIN salaries s
ON s.from_date <= t.to_date
AND s.to_date >= t.from_date;
+----------------+--------+------------+------------+
| name | salary | from_date | to_date |
+----------------+--------+------------+------------+
| Georgi Fecello | 60117 | 1986-06-26 | 1987-06-26 |
| Georgi Fecello | 62102 | 1987-06-26 | 1988-06-25 |
| Georgi Fecello | 66074 | 1988-06-25 | 1989-06-25 |
| Georgi Fecello | 66596 | 1989-06-25 | 1990-06-25 |
| Georgi Fecello | 66961 | 1990-06-25 | 1991-06-25 |
| Georgi Fecello | 71046 | 1991-06-25 | 1992-06-24 |
+----------------+--------+------------+------------+
答案 3 :(得分：0)
您可以分隔所有日期，并检查该表中的日期是：
DECLARE @employees TABLE
(
emp_no INT,
name NVARCHAR(MAX),
title NVARCHAR(MAX),
from_date DATE,
to_date DATE
)
DECLARE @salaries TABLE
(
emp_no INT,
salary INT,
from_date DATE,
to_date DATE
)
INSERT INTO @employees VALUES
(1,'Margareta','Manager','1985-01-01','1991-10-01'),
(2,'Georgi Facello','Employee','1986-06-26','2002-06-22')
INSERT INTO @salaries VALUES
(2,60117,'1986-06-26','1987-06-26'),
(2,62102,'1987-06-26','1988-06-25'),
(2,66074,'1988-06-25','1989-06-25'),
(2,66596,'1989-06-25','1990-06-25'),
(2,66961,'1990-06-25','1991-06-25'),
(2,71046,'1991-06-25','1992-06-24'),
(2,74333,'1992-06-24','1993-06-24'),
(2,75286,'1993-06-24','1994-06-24'),
(2,75994,'1994-06-24','1995-06-24'),
(2,76884,'1995-06-24','1996-06-23'),
(2,80013,'1996-06-23','1997-06-23'),
(2,81025,'1997-06-23','1998-06-23'),
(2,81097,'1998-06-23','1999-06-23'),
(2,84917,'1999-06-23','2000-06-22'),
(2,85112,'2000-06-22','2001-06-22'),
(2,85097,'2001-06-22','2002-06-22')
DECLARE @from_date DATE = (SELECT from_date FROM @employees WHERE title = 'Manager' AND name = 'Margareta')
DECLARE @to_date DATE = (SELECT to_date FROM @employees WHERE title = 'Manager' AND name = 'Margareta')
DECLARE @dates TABLE ( singleDate DATE )
INSERT INTO @dates
SELECT TOP (DATEDIFF(DAY, @from_date, @to_date) + 1)
Date = DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY a.object_id) - 1, @from_date)
FROM sys.all_objects a
CROSS JOIN sys.all_objects b
SELECT E.name,S.* FROM @salaries S
LEFT JOIN @employees E ON S.emp_no = E.emp_no
WHERE S.from_date IN (SELECT singleDate FROM @dates)
OR S.to_date IN (SELECT singleDate FROM @dates)
输出：
name emp_no salary from_date to_date
Georgi Facello 2 60117 1986-06-26 1987-06-26
Georgi Facello 2 62102 1987-06-26 1988-06-25
Georgi Facello 2 66074 1988-06-25 1989-06-25
Georgi Facello 2 66596 1989-06-25 1990-06-25
Georgi Facello 2 66961 1990-06-25 1991-06-25
Georgi Facello 2 71046 1991-06-25 1992-06-24