Question

在PostgreSQL中获得准确快速查询以获得最长前缀匹配的最佳方法是什么？

是吗：

A.) select * from table where column in (subselect) ;

B.) select * from table where strpos(column,column2) = 1
    order by length(column2) desc limit 1 ;

C.) select * from table where column ~ column2
    order by length(column2) desc limit 1

我打算在更新中使用。有什么想法吗？

Answer 1

我不知道在PostgreSQL中开箱即用的功能。
recursive CTE将是一个相当优雅的解决方案的关键元素（在PostgreSQL 8.4或更高版本中可用）。

我假设一个表filter来保存过滤字符串：

CREATE TABLE filter (f_id int, string text);

要搜索最长匹配的表tbl：

CREATE TABLE tbl(t_id int, col text);

查询

WITH RECURSIVE
     f AS (SELECT f_id, string, length(string) AS flen FROM filter)
    ,t AS (SELECT t_id, col, length(col) AS tlen FROM tbl)
    ,x AS (
    SELECT t.t_id, f.f_id, t.col, f.string
          ,2 AS match, LEAST(flen, tlen) AS len
    FROM   t
    JOIN   f ON left(t.col, 1) = left(f.string, 1)

    UNION ALL
    SELECT t_id, f_id, col, string, match + 1, len
    FROM   x
    WHERE  left(col, match) = left(string, match)
    AND    match <= len
    )
SELECT DISTINCT
       f_id
      ,string
      ,first_value(col) OVER w AS col
      ,first_value(t_id) OVER w AS t_id
      ,(first_value(match) OVER w -1) AS longest_match
FROM   x
WINDOW w AS (PARTITION BY f_id ORDER BY match DESC)
ORDER  BY 2,1,3,4;

Detailed explanation how the final SELECT works in this related answer.
Working demo on sqlfiddle.

您没有定义从一组同样长的匹配中选择哪个匹配。我从领带中挑选了一个任意的赢家。

我打算在更新中使用。

PostgreSQL 9.1引入了data modifying CTEs，因此您可以直接在UPDATE语句中使用它。

最长的前缀匹配

1 个答案:

查询