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HenryJi529
AlgorithmJourney
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AlgorithmJourney
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LeetCode
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sql
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LeetCode1965.sql
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48 lines (46 loc) · 1.12 KB
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AlgorithmJourney
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sql
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LeetCode1965.sql
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-- 先分别求缺失数据的id再拼接
select employee_id
from (
select e.employee_id
from Employees as e
left join Salaries as s on e.employee_id = s.employee_id
where
salary is null
union
select s.employee_id
from Employees as e
right join Salaries as s on e.employee_id = s.employee_id
where
name is null
) t
order by employee_id;
-- 先拼接再求缺失数据的id
select DISTINCT
employee_id
from (
select e.employee_id, e.name, s.salary
from Employees e
left join Salaries s on e.employee_id = s.employee_id
union all
select s.employee_id, e.name, s.salary
from Employees e
right join Salaries s on e.employee_id = s.employee_id
) t
where
t.name is null
or t.salary is null
order by employee_id;
-- 统计数据次数
select employee_id
FROM (
SELECT employee_id
FROM Employees
UNION ALL
SELECT employee_id
FROM Salaries
) AS t
GROUP BY
employee_id
HAVING
COUNT(*) = 1
ORDER BY employee_id ASC;
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