SQL 中的分层控制范围报告,无需 Oracle CONNECT BY 语法?

2023-12-28

Summary

控制范围是指有多少员工向特定经理报告。直接和间接报告计数应分为各自的总数。 还需要进行其他统计,包括组织中直接和间接报告的许多职位空缺。经理是指有其他职位向其报告的任何职位。 需要从顶部到树中任何位置的报告路径flatten结构。

我发现这个问题经常出现在人力资源报告和数据仓库项目中。我只能在Oracle中解决这个问题。 此报告可以用与其他数据库(例如 SQL Server 或 PostgreSQL)兼容的 (ANSI) SQL 编写吗?

Detail

组织层次结构的视觉表示:

Level 1                                1:3
                                        |
                        ----------------+-----------------------------------
                        |               |               |                  |
Level 2                2:1            13:             10:12               4:2
                        |                               |
               ---------+----------           ----------+----------
               |        |         |           |         |         |
Level 3      12:10     3:        3:         5:10-1    11:11      6:
               |                              |                   |
            ---+---               ------------+------------       |
            |     |               |     |     |     |     |       |
Level 4    7:4   7:9             8:5   8:7   8:6   8:    8:      9:8

树的每个节点或叶子由以下之一表示:

  • position_id:employee_id
  • position_id:employee_id-multi_job_sequence (if multi_job_sequence>0)
  • position_id:(空的)

预期产出

POSITION_ID    POSITION_DESCR         REPORTSTO_POSITION_ID      EMPLOYEE_ID    MULTI_JOB_SEQUENCE      EMPLOYEE_NAME      TREE_LEVEL_NUM      IS_MANAGER     MAX_INCUMBENTS       FILLED_HEAD_COUNT      VACANT_HEAD_COUNT     FILLED_DIRECT_REPORTS     VACANT_DIRECT_REPORTS       FILLED_INDIRECT_REPORTS     VACANT_INDIRECT_REPORTS       EMPLOYEES_UNDER_POSITION        VACANCIES_UNDER_POSITION       REPORTING_PATH_POSITION_ID     REPORTING_PATH_POSITION_DESCR                       REPORTING_PATH_EMPLOYEE        REPORTING_PATH_EMPLOYEE_NAME
1              CEO                    NULL                       3              0                       Jill               1                   1              1                    1                      0                     3                         1                           9                           5                             12                              6                              1                              CEO                                                 3                              Jill
2              Senior Manager         1                          1              0                       Tom                2                   1              1                    1                      0                     1                         2                           2                           0                             3                               2                              1>2                            CEO>Senior Manager                                  3>1                            Jill>Tom
3              West Winger            2                          NULL           NULL                    NULL               3                   0              2                    0                      2                     0                         0                           0                           0                             0                               0                              1>2>3                          CEO>Senior Manager>West Winger                      3>1>(vacant)                   Jill>Tom>(vacant)
4              Executive Assistant    1                          2              0                       Doug               2                   0              1                    1                      0                     0                         0                           0                           0                             0                               0                              1>4                            CEO>Executive Assistant                             3>2                            Jill>Doug
5              Supervisor South       10                         10             1                       Frank              3                   1              1                    1                      0                     3                         2                           0                           0                             3                               2                              1>10>5                         CEO>Senior Manager>Supervisor South                 3>12>10-1                      Jill>Fred>Frank
6              Supervisor East        10                         NULL           NULL                    NULL               3                   1              1                    0                      1                     1                         0                           0                           0                             1                               0                              1>10>6                         CEO>Senior Manager>Supervisor East                  3>12>(vacant)                  Jill>Fred>(vacant)
7              Expert                 12                         4              0                       Olivia             4                   0              2                    2                      0                     0                         0                           0                           0                             0                               0                              1>2>12>7                       CEO>Senior Manager>Supervisor West>Expert           3>1>10>4                       Jill>Tom>Frank>Olivia
7              Expert                 12                         9              0                       David              4                   0              2                    2                      0                     0                         0                           0                           0                             0                               0                              1>2>12>7                       CEO>Senior Manager>Supervisor West>Expert           3>1>10>9                       Jill>Tom>Frank>David
8              Minion                 5                          5              0                       Carol              4                   0              5                    3                      2                     0                         0                           0                           0                             0                               0                              1>10>5>8                       CEO>Senior Manager>Supervisor South>Minion          3>12>10-1>5                    Jill>Fred>Frank>Carol
8              Minion                 5                          6              0                       Mary               4                   0              5                    3                      2                     0                         0                           0                           0                             0                               0                              1>10>5>8                       CEO>Senior Manager>Supervisor South>Minion          3>12>10-1>6                    Jill>Fred>Frank>Mary
8              Minion                 5                          7              0                       Michael            4                   0              5                    3                      2                     0                         0                           0                           0                             0                               0                              1>10>5>8                       CEO>Senior Manager>Supervisor South>Minion          3>12>10-1>7                    Jill>Fred>Frank>Michael
9              Administrator          6                          8              0                       Nigel              4                   0              1                    1                      0                     0                         0                           0                           0                             0                               0                              1>10>6>9                       CEO>Senior Manager>Supervisor East>Administrator    3>12>(vacant)>8                Jill>Fred>(vacant)>Nigel
10             Senior Manager         1                          12             0                       Fred               2                   1              1                    1                      0                     2                         1                           4                           2                             6                               3                              1>10                           CEO>Senior Manager                                  3>12                           Jill>Fred
11             Supervisor South       10                         11             0                       Wilson             3                   0              1                    1                      0                     0                         0                           0                           0                             0                               0                              1>10>11                        CEO>Senior Manager>Supervisor South                 3>12>11                        Jill>Fred>Wilson
12             Supervisor West        2                          10             0                       Frank              3                   1              1                    1                      0                     2                         0                           0                           0                             2                               0                              1>2>12                         CEO>Senior Manager>Supervisor West                  3>1>10                         Jill>Tom>Frank
13             Executive Mid-West     1                          NULL           NULL                    NULL               2                   0              1                    0                      1                     0                         0                           0                           0                             0                               0                              1>13                           CEO>Executive Mid-West                              3>(vacant)                     Jill>(vacant)

技术要求

  1. The reportsto_position_id包含经理的position_id, NULL 表示顶部位置。
  2. The position_id必须始终存在,但可以为空。
  3. 管理者必须有独特的position_id (and max_incumbents=1)以使树正常工作。
  4. 不同子树或不同层次的相似位置也必须有不同的position_id维持报告结构。这是因为reportsto_position_id为树中的每个节点定义。
  5. An employee_id可以存在于多个节点上,表明该员工在组织中拥有多个职位。如果一名员工有 1 份工作,他们的multi_job_sequence0。如果一名员工身兼数职,他们的multi_job_sequence是递增的。
  6. 职位有一个max_incumbents限制允许填补该职位的员工数量。职位空缺没有职位行,但可以计算。
  7. 经理职位可能会空缺,即使员工仍然向该职位汇报。
  8. 如果组织决定通过添加/删除级别或子树来进行重组,则 SQL 代码不应更改。
  9. 这个例子过于简单化了。大型组织可以为职位和员工提供更多级别和选项(例如生效日期或状态)。为了降低复杂性,本示例中的所有员工和职位均处于活动状态。

控制范围报告业务要求

报告必须回答以下在层级组织中常见的问题:

  1. 经理有多少名直接下属(仅比他们低一级的员工数量)?
  2. 经理有多少个间接报告(比他们低一级的员工计数,一直到树的最低级别)?
  3. 这位经理“在其职位下”有多少人(即直接下属+间接下属)?
  4. 有多少经理需要填补团队中的空缺职位(空缺的直接下属)?
  5. 有多少经理的下属经理的团队中有空缺(空缺的间接下属)?
  6. 从顶部到树中每个位置的路径是什么(按名称或 ID):例如CEO>Senior Manager>Supervisor South>Minion, or 1>2>5>8?
  7. 从顶部到树中每个员工的路径是什么,按姓名或 ID(考虑到可能有多个工作的员工):例如Jill>Tom>Frank>Olivia or 3>1>10-1>4?

样本数据

position table

position_id  descr                            reportsto_position_id  max_incumbents
1            CEO                              NULL                   1
2            Senior Manager                   1                      1
3            West Winger                      2                      2
4            Executive Assistant              1                      1
5            Supervisor South                 10                     1
6            Supervisor East                  10                     1
7            Expert                           12                     2
8            Minion                           5                      5
9            Administrator                    6                      1
10           Senior Manager                   1                      1
11           Supervisor South                 10                     1
12           Supervisor West                  2                      1
13           Executive Mid-West               1                      1

job table

employee_id  multi_job_sequence  employee_name  position_id
1            0                   Tom            2
2            0                   Doug           4
3            0                   Jill           1
4            0                   Olivia         7
5            0                   Carol          8
6            0                   Mary           8
7            0                   Michael        8
8            0                   Nigel          9
9            0                   David          7
10           0                   Frank          12
10           1                   Frank          5
11           0                   Wilson         11
12           0                   Fred           10

SQL

-- Position incumbents. One row for each position, employee_id, multi_job_sequence combination.
with cte_incumbents
as
(
    select
    cp.position_id,
    cp.reportsto_position_id,
    cp.max_incumbents,
    cj.employee_id,
    cj.multi_job_sequence
    from position cp
    left join job cj on cj.position_id = cp.position_id
),
-- Incumbents count (filled and vacant) per position
cte_incumbents_count
as
(
    select
    i.reportsto_position_id,
    i.position_id,
    count(to_char(i.employee_id) || '-' || to_char(i.multi_job_sequence)) as filled_count,
    (i.max_incumbents - count(to_char(i.employee_id) || '-' || to_char(i.multi_job_sequence))) as vacant_count,
    i.max_incumbents
    from cte_incumbents i
    where i.employee_id is not null
    group by i.reportsto_position_id,
             i.position_id,
             i.max_incumbents

    UNION ALL

    select
    i.reportsto_position_id,
    i.position_id,
    0 as filled_count,
    (count(*) * i.max_incumbents) as vacant_count,
    i.max_incumbents
    from cte_incumbents i
    where i.employee_id is null
    group by i.reportsto_position_id,
             i.position_id,
             i.max_incumbents
),
-- Count the filled and vacant reports_to positions
cte_reportsto_count
as
(
    select
    i.reportsto_position_id,
    sum(i.filled_count) as filled_count,
    sum(i.vacant_count) as vacant_count,
    sum(i.max_incumbents) as total_incumbents
    from cte_incumbents_count i
    group by i.reportsto_position_id
),
-- Create the organisation tree, based on the reportsto_position_id
cte_reportsto_tree
as
(
    select
    rtt.position_id,
    rtt.employee_id,
    rtt.multi_job_sequence,
    rtt.position_descr,
    rtt.reportsto_position_id,
    rtt.employee_name,
    level as tree_level_num,
    case when connect_by_isleaf = 0 then 1 else 0 end as is_manager,
    rtt.max_incumbents,
    nvl((
        select
        rtc.filled_count
        from cte_reportsto_count rtc
        where rtc.reportsto_position_id = rtt.position_id
    ),0) as filled_direct_reports,
    nvl((
        select
        rtc.vacant_count
        from cte_reportsto_count rtc
        where rtc.reportsto_position_id = rtt.position_id
    ),0) as vacant_direct_reports,
    substr(sys_connect_by_path(rtt.position_id,'>'),2,length(sys_connect_by_path(rtt.position_id,'>'))-1) as reporting_path_position_id,
    substr(sys_connect_by_path(rtt.position_descr,'>'),2,length(sys_connect_by_path(rtt.position_descr,'>'))-1) as reporting_path_position_descr,
    substr(sys_connect_by_path(nvl(case when rtt.employee_id is null then null else case when rtt.multi_job_sequence = 0 then to_char(rtt.employee_id) else rtt.employee_id || '-' || rtt.multi_job_sequence end end,'(vacant)'),'>'),2,length(sys_connect_by_path(nvl(case when rtt.employee_id is null then null else rtt.employee_id || '-' || rtt.multi_job_sequence end,'(vacant)'),'>'))-1) as reporting_path_employee,
    substr(sys_connect_by_path(nvl(rtt.employee_name,'(vacant)'),'>'),2,length(sys_connect_by_path(nvl(rtt.employee_name,'(vacant)'),'>'))-1) as reporting_path_name
    from
    (
        select
        cp.position_id,
        cp.descr as position_descr,
        cp.max_incumbents,
        cp.reportsto_position_id,
        cj.employee_id,
        cj.multi_job_sequence,
        cj.employee_name
        from position cp
        left join job cj on cj.position_id = cp.position_id -- Positions may not be filled
    ) rtt
    connect by prior rtt.position_id = rtt.reportsto_position_id
    start with rtt.reportsto_position_id is null -- Start at the top of the tree
),
-- Create the report detail, traversing the tree (creating subtrees to get the indirect values). This is the tough part!
cte_report_detail
as
(
    select
    soc.position_id,
    soc.position_descr,
    soc.reportsto_position_id,
    soc.employee_id,
    soc.multi_job_sequence,
    soc.employee_name,
    soc.tree_level_num,
    soc.is_manager,
    soc.max_incumbents,
    nvl(
        (
         select
         ic.filled_count
         from cte_incumbents_count ic
         where ic.position_id = soc.position_id
        ),0) as filled_head_count,
    nvl(
        (
         select
         ic.vacant_count
         from cte_incumbents_count ic
         where ic.position_id = soc.position_id
        ),0) as vacant_head_count,
    soc.filled_direct_reports as filled_direct_reports,
    soc.vacant_direct_reports as vacant_direct_reports,
    case when soc.is_manager = 1 then
    -- Get the filled count of all of the positions underneath and subtract the direct reports to arrive at the filled indirect reports count
    (
        select
        sum(
             (
                select
                rtc.filled_count
                from cte_reportsto_count rtc
                where rtc.reportsto_position_id = cp.position_id
             )
           )
        from position cp
        connect by prior cp.position_id = cp.reportsto_position_id
        start with cp.position_id = soc.position_id
    ) - soc.filled_direct_reports else 0 end as filled_indirect_reports,
    -- Get the vacant count of all of the positions underneath and subtract the direct reports to arrive at the vacant indirect reports count
    case when soc.is_manager = 1 then
    (
        select
        sum(
             (
                select
                rtc.vacant_count
                from cte_reportsto_count rtc
                where rtc.reportsto_position_id = cp.position_id
             )
           )
        from position cp
        connect by prior cp.position_id = cp.reportsto_position_id
        start with cp.position_id = soc.position_id
    ) - soc.vacant_direct_reports else 0 end as vacant_indirect_reports,
    to_clob(cast(soc.reporting_path_position_id as varchar2(4000))) as reporting_path_position_id,
    to_clob(cast(soc.reporting_path_position_descr as varchar2(4000))) as reporting_path_position_descr,
    to_clob(cast(soc.reporting_path_employee as varchar2(4000))) as reporting_path_employee,
    to_clob(cast(soc.reporting_path_name as varchar2(4000))) as reporting_path_employee_name
    from cte_reportsto_tree soc
)
-- Final calculations and sort
select
r.position_id,
r.position_descr,
r.reportsto_position_id,
r.employee_id,
r.multi_job_sequence,
r.employee_name,
r.tree_level_num,
r.is_manager,
r.max_incumbents,
r.filled_head_count,
r.vacant_head_count,
r.filled_direct_reports,
r.vacant_direct_reports,
r.filled_indirect_reports,
r.vacant_indirect_reports,
(r.filled_direct_reports + r.filled_indirect_reports) as employees_under_position,
(r.vacant_direct_reports + r.vacant_indirect_reports) as vacancies_under_position,
r.reporting_path_position_id,
r.reporting_path_position_descr,
r.reporting_path_employee,
r.reporting_path_employee_name
from cte_report_detail r
order by r.position_id,
         r.employee_id,
         r.multi_job_sequence;

SQL 小提琴示例 http://sqlfiddle.com/#!4/52637/13


简而言之,答案是肯定的。

标准 SQL:1999 定义了“递归 CTE”(递归公用表表达式),它执行以下操作:CONNECT BY以及更多。它们被设计用于遍历任何类型的图形——层次结构是它们可以处理的子集。

您的查询非常广泛,因此我没有时间仔细检查并用标准 SQL 重写它。

你询问哪些数据库可以做到这一点。嗯,它们目前由以下人员实施:

  • Oracle.
  • DB2。在 Linux/Unix/Windows 中不实现循环检测。 z/OS 中也是如此。
  • PostgreSQL。
  • SQL Server(从 2012 年开始?)。不实现循环检测。
  • MariaDB,自 10.2 起。不实现循环检测。
  • MySQL 从 8.0 开始。不实现循环检测。
  • H2(从 1.4 开始?)。不实现循环检测。
  • 超SQL。
  • 其他数据库...

如果您提供一个较小的示例,我将非常有兴趣使用递归 CTE 重新表述它。

例如,以下递归 CTE(在 Oracle 中)将找到(直接和间接)向职位 = 2 报告的所有员工的子树:

with
x (position_id, descr, reportsto_position_id, max_incumbents, cur_level) as (
  select
    position_id, descr, reportsto_position_id, max_incumbents,
    1
    from position
    where position_id = 2 -- start at position = 2
  union all
  select
    p.position_id, p.descr, p.reportsto_position_id, p.max_incumbents,
    x.cur_level + 1
    from position p
    join x on p.reportsto_position_id = x.position_id
)
select * from x;
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