如何制作订购单？ [英] How to make ORDER SIBLINGS BY?

问题描述

我正在使用PostgreSQL 9.1.6并尝试构建递归SQL。

我想排序类似于SQL Server中的 ORDER SIBLINGS BY

I'm using PostgreSQL 9.1.6 and trying to build recursive SQL.
I want to sort like ORDER SIBLINGS BY in SQL-Server does it.

_{编者注：这可能是指 Oracle ，其中订购单价实际上存在。}

测试表：

create table RECURSIVE_TEST(
  EMP_ID int,
  MANAGER_ID int,
  EMP_NAME varchar(30)
);

insert into recursive_test values
(1  ,0 ,'MANAGER1'),
(2  ,0 ,'MANAGER2'),
(3  ,0 ,'MANAGER3'),
(4  ,0 ,'MANAGER4'),
(5  ,1 ,'emp1'),
(6  ,3 ,'emp2'),
(7  ,4 ,'emp3'),
(8  ,2 ,'emp4'),
(9  ,2 ,'emp5'),
(10 ,3 ,'emp6'),
(11 ,4 ,'emp7'),
(12 ,3 ,'emp8'),
(13 ,4 ,'emp9'),
(14 ,2 ,'emp10'),
(15 ,2 ,'emp11'),
(16 ,1 ,'emp12');

查询：

WITH RECURSIVE T AS
(
    SELECT A.EMP_ID
             ,A.MANAGER_ID
             ,A.EMP_NAME
      FROM RECURSIVE_TEST A
    WHERE MANAGER_ID = 0
    UNION ALL
    SELECT A.EMP_ID
             ,A.MANAGER_ID
             ,A.EMP_NAME
      FROM RECURSIVE_TEST A, T
    WHERE A.MANAGER_ID = T.EMP_ID
)
SELECT * FROM T;

结果：

 emp_id | manager_id | emp_name 
--------+------------+----------
      1 |          0 | MANAGER1
      2 |          0 | MANAGER2
      3 |          0 | MANAGER3
      4 |          0 | MANAGER4
      5 |          1 | emp1
      6 |          3 | emp2
      7 |          4 | emp3
      8 |          2 | emp4
      9 |          2 | emp5
     10 |          3 | emp6
     11 |          4 | emp7
     12 |          3 | emp8
     13 |          4 | emp9
     14 |          2 | emp10
     15 |          2 | emp11
     16 |          1 | emp12

我想对结果集进行如下排序。

I want to sort the result-set up there like below.

 emp_id | manager_id | emp_name 
--------+------------+----------
      1 |          0 | MANAGER1
      5 |          1 | emp1
     16 |          1 | emp12      
      2 |          0 | MANAGER2
      8 |          2 | emp4
      9 |          2 | emp5
     14 |          2 | emp10
     15 |          2 | emp11      
      3 |          0 | MANAGER3
      6 |          3 | emp2
     10 |          3 | emp6
     12 |          3 | emp8     
      4 |          0 | MANAGER4
      7 |          4 | emp3
     11 |          4 | emp7
     13 |          4 | emp9

有什么建议吗？

推荐答案

这将实现您所描述的内容：

This would achieve what you describe:

WITH RECURSIVE t AS (
   SELECT emp_id As top_id
        , emp_id
        , manager_id
        , emp_name
   FROM   recursive_test
   WHERE  manager_id = 0

   UNION ALL
   SELECT t.top_id
        , a.emp_id
        , a.manager_id
        , a.emp_name
   FROM   recursive_test a
   JOIN   t ON a.manager_id = t.emp_id
   )
SELECT emp_id
     , manager_id
     , emp_name
FROM   t
ORDER  BY top_id, emp_id;

似乎您想按 emp_id 中学..

WITH RECURSIVE t AS (
   SELECT ARRAY[emp_id] AS hierarchy
        , emp_id
        , manager_id
        , emp_name
   FROM   recursive_test
   WHERE  manager_id = 0

   UNION ALL
   SELECT t.hierarchy || a.emp_id
        , a.emp_id
        , a.manager_id
        , a.emp_name
   FROM   recursive_test a
   JOIN   t ON a.manager_id = t.emp_id
   )
SELECT emp_id
     , manager_id
     , emp_name
FROM   t
ORDER  BY hierarchy;

此函数以数组的形式收集祖先（包括自身）并对其进行排序，从而实现像

This one collects ancestors (including self) in an array and sorts by it, achieving an ordering like in a table of contents.

按数组排序按预期工作。相关：

Sorting by an array works as expected. Related:

• 如何订购典型的软件发行版本，例如XYZ？

• How to ORDER BY typical software release versions like X.Y.Z?

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