SQL - 根据特定条件查找重复记录答案

【问题标题】：SQL - Finding Duplicate Records based certain criteriaSQL - 根据特定条件查找重复记录
【发布时间】：2022-12-06 21:12:46
【问题描述】：

我在表中有这些记录 - employee_projects

id	employee_id	project_id	status
1	emp1	proj1	VERIFIED
2	emp2	proj2	REJECTED
3	emp1	proj1	VERIFIED
4	emp1	proj3	REJECTED
5	emp2	proj2	REQUIRED
6	emp3	proj4	SUBMITTED
7	emp4	proj5	VERIFIED
8	emp4	proj6	VERIFIED
9	emp3	proj4	REQUIRED

以下是确定重复项的标准：

同一员工ID，同一状态下的同一项目ID（示例：第1行和第3行重复）
相同的员工 ID、相同的项目 ID 但状态不同（示例：第 6 行和第 9 行重复）。重复标准#2 的一个例外是，如果一个项目是必需的，而同一个项目在同一员工下也被拒绝，则这不被视为重复。例如，第 2 行和第 5 行不是重复的。
我有一个关于第一个标准的查询：
```
select
    emp_id,
    proj_id,
    status,
    COUNT(*)
from
    employee_projects
group by
    emp_id,
    proj_id,
    status
having
    COUNT(*) > 1
```
我正在努力构建的是第二个标准的 SQL。

【问题讨论】：

如果对于相同的 emp_id、proj_id，你有 status 的 'REJECTED'、'REQUIRED'、'REJECTED'、'REJECTED'...等，这是否被视为重复？

标签： sql

【解决方案1】：

也许自我加入可以帮助你。

with t (employee_id ,project_id,status)
as
(
select 'emp1',  'proj1' ,   'VERIFIED'
Union all select 'emp2',    'proj2' ,   'REJECTED'
Union all select 'emp1',    'proj1' ,   'VERIFIED'
Union all select 'emp1',    'proj3' ,   'REJECTED'
Union all select 'emp2',    'proj2' ,   'REQUIRED'
Union all select 'emp3',    'proj4' ,   'SUBMITTED'
Union all select 'emp4',    'proj5' ,   'VERIFIED'
Union all select 'emp4',    'proj6' ,   'VERIFIED'
Union all select 'emp3',    'proj4' ,   'REQUIRED'
)
select
    t.employee_id,
    t.project_id,
    t.status,
    '' as status,
    'criteria#1' as SQL
from
    t
group by
    t.employee_id,
    t.project_id,
    t.status
having
    COUNT(*) > 1
union all
SELECT 
    t.employee_id,
    t.project_id,
    t.status,
    a.status,
    'criteria#2' as SQL
FROM
    t
    left join t as a on 
        t.employee_id = a.employee_id and
        t.project_id = a.project_id
where 
    t.status != a.status and
    concat(t.status,a.status) != 'REQUIREDREJECTED' and
    concat(t.status,a.status) != 'REJECTEDREQUIRED'

【讨论】：

【解决方案2】：

尝试以下操作：

select T.emp_id, T.proj_id, T.status, D.dup_cnt
from employee_projects T join
(
  select emp_id, proj_id, count(*) as dup_cnt
  from employee_projects
  group by emp_id, proj_id
  having count(*) > 1 and
    count(distinct case when status in ('REQUIRED', 'REJECTED') then status end) < 2
) D
on T.emp_id = D.emp_id and T.proj_id = D.proj_id
order by T.emp_id, T.proj_id

如果您想将具有状态（'REQUIRED'、'REJECTED'、任何其他状态）的员工视为重复员工，请修改 having 子句，如下所示：

select T.emp_id, T.proj_id, T.status, D.dup_cnt
from employee_projects T join
(
  select emp_id, proj_id, count(*) as dup_cnt
  from employee_projects
  group by emp_id, proj_id
  having count(*) > 1 and
    (count(distinct case when status in ('REQUIRED', 'REJECTED') then status end) < 2 or count(distinct status) > 2)
) D
on T.emp_id = D.emp_id and T.proj_id = D.proj_id
order by T.emp_id, T.proj_id

见demo。

【讨论】：