带连接的PostgreSQL反向查询

根据下面的模式，我试图让第一部门的所有员工都知道他们在某个特定日期缺席（如果有）的理由，如下所示

-------------------------------------------------------------------
id     | department |      full_name    |  excuse      | date
-------+------------+-------------------+--------------+-----------
    10 |      Sales | John Doe          | NULL         | 2019-01-20
    20 |      Sales | James Brown       | NULL         | 2019-01-20
    30 |      Sales | Adam Sanders      | Fatigued     | 2019-01-20
    70 |      Sales | Amber Philips     | NULL         | 2019-01-20

我已经尝试了下面的查询，但是我没有得到正确的答案

SELECT DISTINCT(emp.id) as employee_id, 
emp.full_name, ab.excuse
dept.name AS department 
FROM employees emp
INNER JOIN departmens dept ON emp.department_id = dept.id
LEFT JOIN absence ab ON ab.employee_id = emp.id
WHERE emp.department_id IN (1)  
AND emp.id NOT IN (
    SELECT att.employee_id FROM attendance att 
    INNER JOIN employees emp ON att.employee_id = emp.id
    WHERE emp.department_id IN (1)
    AND att.date BETWEEN SYMMETRIC '2019-01-20' AND '2019-01-20'
)

员工

-----------------------------------------
id     | department |      full_name  
-------+------------+--------------------
    10 |          1 | John Doe           
    20 |          1 | James Brown        
    30 |          1 | Adam Sanders   
    40 |          3 | Craig Wallace
    50 |          3 | Lina Goode
    60 |          6 | Jenny Rosewood 
    70 |          1 | Amber Philips   

出席人数

----------------------------------------
id     | employee_id |  attendance_date
-------+-------------+------------------
     1 |          10 | 2019-01-20
     2 |          20 | 2019-01-20

-------------------------------------------------------
id     | employee_id |       excuse       | date
-------+-------------+--------------------+------------
     1 |          30 | Fatigued           | 2019-01-20
     2 |          20 | Feeling Sick       | 2019-01-18
     3 |          10 | Out Of town        | 2019-01-15

缺席

----------------------------------------
id     | employee_id |  attendance_date
-------+-------------+------------------
     1 |          10 | 2019-01-20
     2 |          20 | 2019-01-20

-------------------------------------------------------
id     | employee_id |       excuse       | date
-------+-------------+--------------------+------------
     1 |          30 | Fatigued           | 2019-01-20
     2 |          20 | Feeling Sick       | 2019-01-18
     3 |          10 | Out Of town        | 2019-01-15

---

根据所需的

结果

，您不需要

子查询

。您的

左连接就足够了

select distinct(emp.id) as employee_id
    , dept.name
    , emp.full_name
    , ab.excuse
    , ab.date
from employees emp
inner join departmens dept on emp.department_id = dept.id
left join absence ab on ab.employee_id = emp.id and ab.date =  '2019-01-20' 
where emp.department_id in (1)  

根据所需的
结果
，您不需要
子查询
。您的
左连接就足够了

select distinct(emp.id) as employee_id
    , dept.name
    , emp.full_name
    , ab.excuse
    , ab.date
from employees emp
inner join departmens dept on emp.department_id = dept.id
left join absence ab on ab.employee_id = emp.id and ab.date =  '2019-01-20' 
where emp.department_id in (1)