pg distinct 改写递归优化（德哥的思路）

德哥的优化思路巨牛逼，这种递归思维真的太吊了，我目前就缺递归思路。

下面SQL1000W行数据，列的选择性很低，只有两个值（’1’和’11’）都是字符串类型，‘1’只有一条数据，’11’有9999999行数据。

慢SQL：

select distinct col from tt;

                                                      QUERY PLAN                                                      
----------------------------------------------------------------------------------------------------------------------
 HashAggregate  (cost=169247.11..169247.12 rows=1 width=3) (actual time=5082.733..5082.735 rows=2 loops=1)
   Group Key: col
   ->  Seq Scan on tt  (cost=0.00..144247.29 rows=9999929 width=3) (actual time=0.005..275.906 rows=10000000 loops=1)
 Planning Time: 0.365 ms
 Execution Time: 5082.772 ms
(5 行记录)

CTE递归优化：

WITH RECURSIVE t AS (
   (SELECT col FROM tt ORDER BY col LIMIT 1)  
   UNION ALL
   SELECT (SELECT col FROM tt WHERE col > t.col ORDER BY col LIMIT 1)
   FROM t
   WHERE t.col IS NOT NULL
   )
SELECT col FROM t WHERE col IS NOT NULL;

                                                                        QUERY PLAN                                                                        
----------------------------------------------------------------------------------------------------------------------------------------------------------
 CTE Scan on t  (cost=50.84..52.86 rows=100 width=38) (actual time=0.024..0.079 rows=2 loops=1)
   Filter: (col IS NOT NULL)
   Rows Removed by Filter: 1
   CTE t
     ->  Recursive Union  (cost=0.43..50.84 rows=101 width=38) (actual time=0.022..0.076 rows=3 loops=1)
           ->  Limit  (cost=0.43..0.46 rows=1 width=3) (actual time=0.021..0.021 rows=1 loops=1)
                 ->  Index Only Scan using idx_1_2_tt on tt tt_1  (cost=0.43..260443.37 rows=9999929 width=3) (actual time=0.020..0.020 rows=1 loops=1)
                       Heap Fetches: 0
           ->  WorkTable Scan on t t_1  (cost=0.00..4.84 rows=10 width=38) (actual time=0.017..0.017 rows=1 loops=3)
                 Filter: (col IS NOT NULL)
                 Rows Removed by Filter: 0
                 SubPlan 1
                   ->  Limit  (cost=0.43..0.46 rows=1 width=3) (actual time=0.024..0.024 rows=0 loops=2)
                         ->  Index Only Scan using idx_1_2_tt on tt  (cost=0.43..95149.36 rows=3333310 width=3) (actual time=0.024..0.024 rows=0 loops=2)
                               Index Cond: (col > (t_1.col)::text)
                               Heap Fetches: 0
 Planning Time: 0.096 ms
 Execution Time: 0.096 ms
(18 行记录)

里面的逻辑是：

(SELECT col FROM tt ORDER BY col LIMIT 1)

　　根节点通过order by 升序 找到最小的一条数据作为起点。

递归查询:

SELECT (SELECT col FROM tt WHERE col > t.col ORDER BY col LIMIT 1)
FROM t
WHERE t.col IS NOT NULL

　　在第一次迭代中，CTE t 包含值’1’。这个查询将在tt表中寻找col大于’1’的最小值。在数据集中，这将是’11’。

　　在第二次迭代，CTE t 将包含’11’。此时，查询将尝试找到大于’11’的最小值，但没有这样的值，所以返回NULL。

递归结束:
　　当递归查询返回NULL时，递归结束。这时，CTE t 将包含’1’和’11’，返回和distinct 一样逻辑的数据。

理解了整个逻辑后我都吓尿了，就一道算法题，确实要跟巨佬学习才行，加深递归思维。