Question

我正在尝试通过links.name列实现一个简单的搜索功能。我想将关键字与名称值进行匹配，并希望在结果顶部获得具有更多匹配项的名称。我想要/尝试做的是

declare matchings integer := 0
select if "keyword1" ~* name then matchings := matchings + 1 end if
       if "keyword2" ~* name then matchings := matchings + 1 end if
       ... so on for all keywords given ..
       as matchings_count from links order by matchings_count;

这样做的正确语法是什么？不必担心性能，因为链接仅包含1200行，并且增加的行数不会超过1500。任何输入将不胜感激。预先感谢。

Answer 1

完成任务的最简单方法是

with t as(
 select 'key1' as k
 union all select 'key2' as k
 union all select 'key3' as k
)

select count(*) from t
where k ~* '(key1)|(key2)'
;

sqlfiddle

如果您只想计算匹配数

select count(*) from t
where k ~* 'key1'
union all select count(*) from t
where k ~* 'key2'
...
;

Answer 2

在Postgres中，您可以将布尔值转换为数字并将其加起来：

select l.*,
       ( ("keyword1" ~* name)::int +
         ("keyword2" ~* name)::int +
         . . .
       ) as num_matches         
from links l
order by num_matches desc;

PostgreSQL有条件地为每一行生成值

2 个答案: