Postgresql 是否有 postgres CLOSEST 运算符

发布于05月24日

我要找的是这样的东西，比如:

| id | number |
|  1 |     .7 |
|  2 |   1.25 |
|  3 |   1.01 |
|  4 |    3.0 |

查询SELECT * FROM my_table WHEREnumberCLOSEST(1)将返回第3行.我只在乎数字.现在，我有一个过程，它只是在每一行上循环并进行比较，但我认为信息应该可以从b树索引中获得，所以这可能是一个内置的，但我找不到任何文档表明它可以.

推荐答案

我可能有点不懂语法，但这个参数化查询(以原始问题的"1"为例)应该运行得很快，基本上是2个B树查找(假设数字已被索引).

SELECT * FROM
(
  (SELECT id, number FROM t WHERE number >= ? ORDER BY number LIMIT 1) AS above
  UNION ALL
  (SELECT id, number FROM t WHERE number < ? ORDER BY number DESC LIMIT 1) as below
) 
ORDER BY abs(?-number) LIMIT 1;

这个查询计划包含~5e5行(索引为number)的表，如下所示:

psql => explain select * from (
        (SELECT id, number FROM t WHERE number >= 1 order by number limit 1) 
        union all
        (select id, number from t where number < 1 order by number desc limit 1)
) as make_postgresql_happy 
order by abs (1 - number) 
limit 1;
                                                  QUERY PLAN
--------------------------------------------------------------------------------------------------------------
 Limit  (cost=0.24..0.24 rows=1 width=12)
   ->  Sort  (cost=0.24..0.24 rows=2 width=12)
         Sort Key: (abs((1::double precision - public.t.number)))
         ->  Result  (cost=0.00..0.23 rows=2 width=12)
               ->  Append  (cost=0.00..0.22 rows=2 width=12)
                     ->  Limit  (cost=0.00..0.06 rows=1 width=12)
                           ->  Index Scan using idx_t on t  (cost=0.00..15046.74 rows=255683 width=12)
                                 Index Cond: (number >= 1::double precision)
                     ->  Limit  (cost=0.00..0.14 rows=1 width=12)
                           ->  Index Scan Backward using idx_t on t  (cost=0.00..9053.67 rows=66136 width=12)
                                 Index Cond: (number < 1::double precision)
(11 rows)