BigQuery 中的递归/分层查询 [英] Recursive/hierarchical query in BigQuery

问题描述

我有一个递归/层次结构问题，我想在 BigQuery 中解决这个问题.

I have a recursion/hierarchical problem that I'm trying to figure out in BigQuery.

我有一个员工列表，每个员工都有一个经理 ID.我需要能够输入一个 Employee_ID 并返回他们下面的每个人的数组.

I have a list of employees and each employee has a manager ID. I need to be able to enter a single Employee_ID and return an array of every person beneath them.

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);

我想要的输出类似于:

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;

我有以下工作，但它看起来非常丑陋/不方便，并且不支持无限级别:

I have got the following working but it seems very ugly/inconvenient and doesn't support unlimited levels:

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

有没有更好的方法来做到这一点?BigQuery 中是否可以使用递归方法?

Is there a better way to do this? Is a recursive approach possible in BigQuery?

推荐答案

以下为 BigQuery Standard SQL

Below is for 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;  

结果

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    

如果您需要将报告作为数组 - 将上面脚本中的最后一条语句替换为下面

In case if you need Reports as array - replace last statement in above script with below

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;  

注意:取决于层次结构中的嵌套级别 - 您可能需要在 OR run_away_stop > 中调整 1010

Note: depends on level of nesting in your hierarchy - you might need to adjust 10 in OR run_away_stop > 10

这篇关于BigQuery 中的递归/分层查询的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！