3

我有一张这样的桌子：

     a    |  user_id
----------+-------------
  0.1133  |  2312882332
  4.3293  |  7876123213
  3.1133  |  2312332332
  1.3293  |  7876543213
  0.0033  |  2312222332
  5.3293  |  5344343213
  3.2133  |  4122331112
  2.3293  |  9999942333

我想找到一个特定的行 -1.3293 | 7876543213例如 - 并选择最近的 4 行。2 上，2 下，如果可能的话。
排序顺序为 ORDER BY a ASC。

在这种情况下，我会得到：

  0.0033  |  2312222332
  0.1133  |  2312882332
  2.3293  |  9999942333
  3.1133  |  2312332332

如何使用 PostgreSQL 实现这一点？（顺便说一句，我正在使用 PHP。）

PS：对于最后一行或第一行，最近的行将是上方 4 或下方 4。

4

3 回答 3

6

测试用例：

CREATE TEMP TABLE tbl(a float, user_id bigint);
INSERT INTO tbl VALUES
 (0.1133, 2312882332)
,(4.3293, 7876123213)
,(3.1133, 2312332332)
,(1.3293, 7876543213)
,(0.0033, 2312222332)
,(5.3293, 5344343213)
,(3.2133, 4122331112)
,(2.3293, 9999942333);

询问：

WITH x AS (
    SELECT a
          ,user_id
          ,row_number() OVER (ORDER BY a, user_id) AS rn
    FROM   tbl
    ), y AS (
    SELECT rn, LEAST(rn - 3, (SELECT max(rn) - 5 FROM x)) AS min_rn
    FROM   x
    WHERE  (a, user_id) = (1.3293, 7876543213)
    )
SELECT *
FROM   x, y
WHERE  x.rn  > y.min_rn
AND    x.rn <> y.rn
ORDER  BY x.a, x.user_id
LIMIT  4;

返回问题中描述的结果。假设这(a, user_id)是独一无二的。

目前尚不清楚是否a应该是唯一的。这就是为什么我user_id另外排序以打破关系。这也是我使用窗口函数row_number()的原因，而不是 rank()为此。row_number()在任何情况下都是正确的工具。我们想要 4 行。rank()如果排序顺序中有对等点，则会给出未定义的行数。

只要表中至少有 5 行，这总是返回 4 行。接近第一行/最后一行，返回前/后4行。在所有其他情况下，之前/之后的两行。条件行本身被排除在外。

提高性能

这是@Tim Landscheidt 发布的改进版本。如果您喜欢索引的想法，请投票支持他的答案。不要打扰小桌子。但会提高大表的性能- 只要您有合适的索引。最好的选择是在(a, user_id).

WITH params(_a, _user_id) AS (SELECT 5.3293, 5344343213) -- enter params once
    ,x AS  (
    (
    SELECT a
          ,user_id
          ,row_number() OVER (ORDER BY a DESC, user_id DESC) AS rn
    FROM   tbl, params p
    WHERE  a < p._a
       OR  a = p._a AND user_id < p._user_id -- a is not defined unique
    ORDER  BY a DESC, user_id DESC
    LIMIT  5  -- 4 + 1: including central row
    )
    UNION ALL -- UNION right away, trim one query level
    (
    SELECT a
          ,user_id
          ,row_number() OVER (ORDER BY a ASC, user_id ASC) AS rn
    FROM   tbl, params p
    WHERE  a > p._a
       OR  a = p._a AND user_id > p._user_id
    ORDER  BY a ASC, user_id ASC
    LIMIT  5
    )
    )
    , y AS (
    SELECT a, user_id
    FROM   x, params p
    WHERE (a, user_id) <> (p._a, p._user_id) -- exclude central row
    ORDER  BY rn  -- no need to ORDER BY a
    LIMIT  4
    )
SELECT *
FROM   y
ORDER  BY a, user_id   -- ORDER result as requested

@Tim 版本的主要区别：

  • 根据问题(a, user_id)形式的搜索条件，不只是a. 这会以微妙不同的方式改变窗口框架ORDER BY和WHERE子句。

  • UNION马上，不需要额外的查询级别。您需要在两个 UNION 查询周围加上括号以允许单个ORDER BY.

  • 根据要求对结果进行排序。需要另一个查询级别（几乎没有任何成本）。

  • 由于参数在多个地方使用，我将输入集中在一个领先的 CTE 中。
    对于重复使用，您可以几乎“按原样”将此查询包装到 SQL 或 plpgsql 函数中。

于 2012-03-24T14:29:07.870 回答
2

还有一个：

WITH prec_rows AS
  (SELECT a,
          user_id,
          ROW_NUMBER() OVER (ORDER BY a DESC) AS rn
   FROM tbl
   WHERE a < 1.3293
   ORDER BY a DESC LIMIT 4),
     succ_rows AS
  (SELECT a,
          user_id,
          ROW_NUMBER() OVER (ORDER BY a ASC) AS rn
   FROM tbl
   WHERE a > 1.3293
   ORDER BY a ASC LIMIT 4)
SELECT a, user_id
FROM
  (SELECT a,
          user_id,
          rn
   FROM prec_rows
   UNION ALL SELECT a,
                    user_id,
                    rn
   FROM succ_rows) AS s
ORDER BY rn, a LIMIT 4;

AFAIRWITH将实例化一个内存表，因此该解决方案的重点是尽可能限制其大小（在本例中为八行）。

于 2012-03-25T01:56:34.100 回答
0 
set search_path='tmp';

DROP TABLE lutser;
CREATE TABLE lutser
        ( val float
        , num bigint
        );
INSERT INTO lutser(val, num)
VALUES ( 0.1133  ,  2312882332  )
      ,( 4.3293  ,  7876123213  )
      ,( 3.1133  ,  2312332332  )
      ,( 1.3293  ,  7876543213  )
      ,( 0.0033  ,  2312222332  )
      ,( 5.3293  ,  5344343213  )
      ,( 3.2133  ,  4122331112  )
      ,( 2.3293  ,  9999942333  )
        ;

WITH ranked_lutsers AS (
        SELECT val, num
        ,rank() OVER (ORDER BY val) AS rnk
        FROM lutser
        )
SELECT that.val, that.num
        , (that.rnk-this.rnk) AS relrnk
FROM ranked_lutsers that
JOIN ranked_lutsers this ON (that.rnk BETWEEN this.rnk-2 AND this.rnk+2)
WHERE this.val = 1.3293
        ;

结果：

DROP TABLE
CREATE TABLE
INSERT 0 8
  val   |    num     | relrnk 
--------+------------+--------
 0.0033 | 2312222332 |     -2
 0.1133 | 2312882332 |     -1
 1.3293 | 7876543213 |      0
 2.3293 | 9999942333 |      1
 3.1133 | 2312332332 |      2
(5 rows)

正如 Erwin 指出的那样，输出中不需要中间行。此外，应该使用 row_number() 而不是 rank()。

WITH ranked_lutsers AS (
        SELECT val, num
        -- ,rank() OVER (ORDER BY val) AS rnk
        , row_number() OVER (ORDER BY val, num) AS rnk
        FROM lutser
) SELECT that.val, that.num
        , (that.rnk-this.rnk) AS relrnk
FROM ranked_lutsers that
JOIN ranked_lutsers this ON (that.rnk BETWEEN this.rnk-2 AND this.rnk+2 )
WHERE this.val = 1.3293
AND that.rnk <> this.rnk
        ;

结果2：

  val   |    num     | relrnk 
--------+------------+--------
 0.0033 | 2312222332 |     -2
 0.1133 | 2312882332 |     -1
 2.3293 | 9999942333 |      1
 3.1133 | 2312332332 |      2
(4 rows)

UPDATE2：总是选择四个，即使我们在列表的顶部或底部。这使得查询有点难看。（但不像欧文那样丑;-)

WITH ranked_lutsers AS (
        SELECT val, num
        -- ,rank() OVER (ORDER BY val) AS rnk
        , row_number() OVER (ORDER BY val, num) AS rnk
        FROM lutser
) SELECT that.val, that.num
        , ABS(that.rnk-this.rnk) AS srtrnk
        , (that.rnk-this.rnk) AS relrnk
FROM ranked_lutsers that
JOIN ranked_lutsers this ON (that.rnk BETWEEN this.rnk-4 AND this.rnk+4 )
-- WHERE this.val = 1.3293
WHERE this.val = 0.1133
AND that.rnk <> this.rnk
ORDER BY srtrnk ASC
LIMIT 4
        ;

输出：

  val   |    num     | srtrnk | relrnk 
--------+------------+--------+--------
 0.0033 | 2312222332 |      1 |     -1
 1.3293 | 7876543213 |      1 |      1
 2.3293 | 9999942333 |      2 |      2
 3.1133 | 2312332332 |      3 |      3
(4 rows)

更新：具有嵌套 CTE 的版本（具有外连接！！！）。为了方便起见，我在表中添加了一个主键，无论如何，恕我直言，这听起来是个好主意。

WITH distance AS (
        WITH ranked_lutsers AS (
        SELECT id
        , row_number() OVER (ORDER BY val, num) AS rnk
        FROM lutser
        ) SELECT l0.id AS one
        ,l1.id AS two
        , ABS(l1.rnk-l0.rnk) AS dist
        -- Warning: Cartesian product below
        FROM ranked_lutsers l0
        , ranked_lutsers l1 WHERE l0.id <> l1.id

        )
SELECT lu.*
FROM lutser lu
JOIN distance di
ON lu.id = di.two
WHERE di.one= 1
ORDER by di.dist
LIMIT 4 
        ;
于 2012-03-24T14:29:56.323 回答