Рекурсивный / иерархический запрос в BigQuery

Question

У меня есть проблема с рекурсией / иерархией, которую я пытаюсь выяснить в BigQuery.

У меня есть список сотрудников, и у каждого сотрудника есть идентификатор менеджера. Мне нужно иметь возможность ввести один Employee_ID и вернуть массив каждого человека под ними.

CREATE TABLE p_RLS.testHeirarchy
 (
   Employee_ID INT64,
   Employee_Name STRING,
   Position STRING,
   Line_Manager_ID INT64
 );

INSERT INTO p_RLS.testHeirarchy (Employee_ID, Employee_Name, Position, Line_Manager_ID)
VALUES(1,'Joe','Worker',11),
      (2,'James','Worker',11),
      (3,'Jack','Worker',11),
      (4,'Jill','Worker',12),
      (5,'Jan','Worker',12),
      (6,'Jacquie','Worker',13),
      (7,'Joaquin','Worker',14),
      (8,'Jeremy','Worker',14),
      (9,'Jade','Worker',15),
      (10,'Jocelyn','Worker',15),
      (11, 'Bob', 'Store Manager',16),
      (12, 'Bill', 'Store Manager',16),
      (13, 'Barb', 'Store Manager',16),
      (14, 'Ben', 'Store Manager',17),
      (15, 'Burt', 'Store Manager',17),
      (16, 'Sally','Group Manager',18),
      (17, 'Sam','Group Manager',19),
      (18, 'Anna', 'Ops Manager',20),
      (19, 'Amy', 'Ops Manager',20),
      (20, 'Zoe', 'State Manager', NULL);

Мой желаемый результат будет выглядеть так:

SELECT 20 as Employee_ID, [19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1] as Reports;
SELECT 11 as Employee_ID, [3,2,1] as Reports;
SELECT 1 as Employee_ID, [] as Reports;

У меня есть следующие рабочие но это кажется очень уродливым / неудобным и не поддерживает неограниченные уровни:

WITH test as (
SELECT L0.Employee_ID, L0.Employee_Name, L0.Position, L0.Line_Manager_ID,
ARRAY_AGG(DISTINCT L1.Employee_ID IGNORE NULLS) as Lvl1, 
ARRAY_AGG(DISTINCT L2.Employee_ID IGNORE NULLS) as Lvl2, 
ARRAY_AGG(DISTINCT L3.Employee_ID IGNORE NULLS) as Lvl3, 
ARRAY_AGG(DISTINCT L4.Employee_ID IGNORE NULLS) as Lvl4, 
ARRAY_AGG(DISTINCT L5.Employee_ID IGNORE NULLS) as Lvl5, 
ARRAY_AGG(DISTINCT L6.Employee_ID IGNORE NULLS) as Lvl6, 
ARRAY_AGG(DISTINCT L7.Employee_ID IGNORE NULLS) as Lvl7
FROM p_RLS.testHeirarchy as L0
LEFT OUTER JOIN p_RLS.testHeirarchy L1 ON L0.Employee_ID = L1.Line_Manager_ID
LEFT OUTER JOIN p_RLS.testHeirarchy L2 ON L1.Employee_ID = L2.Line_Manager_ID
LEFT OUTER JOIN p_RLS.testHeirarchy L3 ON L2.Employee_ID = L3.Line_Manager_ID
LEFT OUTER JOIN p_RLS.testHeirarchy L4 ON L3.Employee_ID = L4.Line_Manager_ID
LEFT OUTER JOIN p_RLS.testHeirarchy L5 ON L4.Employee_ID = L5.Line_Manager_ID
LEFT OUTER JOIN p_RLS.testHeirarchy L6 ON L5.Employee_ID = L6.Line_Manager_ID
LEFT OUTER JOIN p_RLS.testHeirarchy L7 ON L6.Employee_ID = L7.Line_Manager_ID
WHERE L0.Employee_ID = 16
GROUP BY 1,2,3,4)

SELECT
Employee_ID, ARRAY_CONCAT(
    IFNULL(Lvl1,[]),
    IFNULL(Lvl2,[]),
    IFNULL(Lvl3,[]),
    IFNULL(Lvl4,[]),
    IFNULL(Lvl5,[]),
    IFNULL(Lvl6,[]),
    IFNULL(Lvl7,[])) as All_reports
FROM test

Есть ли лучший способ сделать это? Возможен ли рекурсивный подход в BigQuery?

Answer 1

Ниже для BigQuery Standard SQL

DECLARE rows_count, run_away_stop INT64 DEFAULT 0;

CREATE TEMP TABLE initialData AS WITH input AS (
  SELECT 1 Employee_ID,'Joe' Employee_Name,'Worker' Position,11 Line_Manager_ID UNION ALL
  SELECT 2,'James','Worker',11 UNION ALL
  SELECT 3,'Jack','Worker',11 UNION ALL
  SELECT 4,'Jill','Worker',12 UNION ALL
  SELECT 5,'Jan','Worker',12 UNION ALL
  SELECT 6,'Jacquie','Worker',13 UNION ALL
  SELECT 7,'Joaquin','Worker',14 UNION ALL
  SELECT 8,'Jeremy','Worker',14 UNION ALL
  SELECT 9,'Jade','Worker',15 UNION ALL
  SELECT 10,'Jocelyn','Worker',15 UNION ALL
  SELECT 11, 'Bob', 'Store Manager',16 UNION ALL
  SELECT 12, 'Bill', 'Store Manager',16 UNION ALL
  SELECT 13, 'Barb', 'Store Manager',16 UNION ALL
  SELECT 14, 'Ben', 'Store Manager',17 UNION ALL
  SELECT 15, 'Burt', 'Store Manager',17 UNION ALL
  SELECT 16, 'Sally','Group Manager',18 UNION ALL
  SELECT 17, 'Sam','Group Manager',19 UNION ALL
  SELECT 18, 'Anna', 'Ops Manager',20 UNION ALL
  SELECT 19, 'Amy', 'Ops Manager',20 UNION ALL
  SELECT 20, 'Zoe', 'State Manager', NULL 
)
SELECT * FROM input;

CREATE TEMP TABLE ttt AS 
SELECT Line_Manager_ID, ARRAY_AGG(Employee_ID) Reports FROM initialData WHERE NOT Line_Manager_ID IS NULL GROUP BY Line_Manager_ID;

LOOP
  SET (run_away_stop, rows_count) = (SELECT AS STRUCT run_away_stop + 1, COUNT(1) FROM ttt);

  CREATE OR REPLACE TEMP TABLE ttt1 AS
    SELECT Line_Manager_ID, ARRAY(SELECT DISTINCT Employee_ID FROM UNNEST(Reports) Employee_ID ORDER BY Employee_ID DESC) Reports
    FROM (
      SELECT Line_Manager_ID, ARRAY_CONCAT_AGG(Reports) Reports    
      FROM (
        SELECT t2.Line_Manager_ID, ARRAY_CONCAT(t1.Reports, t2.Reports) Reports 
        FROM ttt t1, ttt t2 
        WHERE (SELECT COUNTIF(t1.Line_Manager_ID = Employee_ID) FROM UNNEST(t2.Reports) Employee_ID) > 0
      ) GROUP BY Line_Manager_ID
    );

  CREATE OR REPLACE TEMP TABLE ttt AS
    SELECT * FROM ttt1 UNION ALL
    SELECT * FROM ttt WHERE NOT Line_Manager_ID IN (SELECT Line_Manager_ID FROM ttt1);

  IF (rows_count = (SELECT COUNT(1) FROM ttt) AND run_away_stop > 1) OR run_away_stop > 10 THEN BREAK; END IF;
END LOOP;

SELECT Employee_ID, 
  (
    SELECT STRING_AGG(CAST(Employee_ID AS STRING), ',' ORDER BY Employee_ID DESC)
    FROM ttt.Reports Employee_ID
  ) Reports_as_list
FROM (SELECT DISTINCT Employee_ID FROM initialData) d
LEFT JOIN ttt ON Employee_ID = Line_Manager_ID
ORDER BY Employee_ID DESC;  

с результатом

Row Employee_ID Reports_as_list  
1   20          19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1  
2   19          17,15,14,10,9,8,7    
3   18          16,13,12,11,6,5,4,3,2,1  
4   17          15,14,10,9,8,7   
5   16          13,12,11,6,5,4,3,2,1     
6   15          10,9     
7   14          8,7  
8   13          6    
9   12          5,4  
10  11          3,2,1    
11  10          null     
12  9           null     
13  8           null     
14  7           null     
15  6           null     
16  5           null     
17  4           null     
18  3           null     
19  2           null     
20  1           null    

В случае, если вам нужен отчет в виде массива - замените последний оператор в приведенном выше скрипте на

SELECT Employee_ID, Reports Reports_as_array
FROM (SELECT DISTINCT Employee_ID FROM initialData) d
LEFT JOIN ttt ON Employee_ID = Line_Manager_ID
ORDER BY Employee_ID DESC;  

Примечание: зависит от уровня вложенности в вашей иерархии - вам может потребоваться настроить 10 в OR run_away_stop > 10

Answer 2

На вопрос: «Возможен ли рекурсивный подход в BigQuery?»

Да!

Теперь, когда BigQuery поддерживает сценарии и циклы, я решил некоторые рекурсивные проблемы из Advent of Code с помощью BigQuery. :

• https://towardsdatascience.com/advent-of-code-sql-bigquery-31e6a04964d4

CREATE TEMP TABLE planets AS SELECT 'YOU' planet;

LOOP
  SET steps = steps+1
  ;
  CREATE OR REPLACE TEMP TABLE planets AS
  SELECT DISTINCT planet
  FROM (
    SELECT origin planet FROM t1 WHERE dest IN (SELECT planet FROM planets)
    UNION ALL
    SELECT dest planet FROM t1 WHERE origin IN (SELECT planet FROM planets)
  )
  ;
  IF 'SAN' IN (SELECT * FROM planets )
      THEN LEAVE;
  END IF;
END LOOP
;
SELECT steps-2

Я бы использовал аналогичный подход для навигации по графику и аннотирования всех родительских отношений.

Скоро: я напишу в блоге статью о специфике обхода дерева, чтобы все оказались под x. Но этот код поможет вам в то же время.