I have a table with the following content :
我有一张包含以下内容的表格:
ID | Name | Alias
ID |名称|别号
1 | William | Will,Willo,Wolli
1 |威廉|请问,Willo,伍利
I would like to return the row ID if the levenshtein distance (or metaphone, it does not matter) of a user-supplied string is lower than a defined threshold for the user name or any of the known aliases.
如果用户提供的字符串的levenshtein距离(或metaphone,无关紧要)低于用户名或任何已知别名的定义阈值,我想返回行ID。
I know that a possible solution is to use an additional table linking user IDs with user aliases, although I'd like to avoid it if possible.
我知道一个可能的解决方案是使用一个将用户ID与用户别名相关联的附加表,尽管如果可能的话我想避免使用它。
2 个解决方案
#1
2
What you need is string split/explode. It could be done like this:
你需要的是字符串拆分/爆炸。可以这样做:
SELECT DISTINCT u.id FROM users AS u LEFT JOIN
(SELECT u.id,unnest(string_to_array(u.alias, ',')) AS ALIAS FROM users AS u) AS q
ON u.id=q.id
WHERE levenshtein(u.name,'Jill')<3
OR levenshtein(q.ALIAS,'Jill')<3;
#2
2
As usual, there is more than one solution:
像往常一样,有多个解决方案:
select u.id
from users u
where 3 >
any
(
select levenshtein ( 'Willey'::text, a )
from regexp_split_to_table
(
concat_ws ( ',' , u.name::text , u.alias::text )
, ','
) as a
)
#1
2
What you need is string split/explode. It could be done like this:
你需要的是字符串拆分/爆炸。可以这样做:
SELECT DISTINCT u.id FROM users AS u LEFT JOIN
(SELECT u.id,unnest(string_to_array(u.alias, ',')) AS ALIAS FROM users AS u) AS q
ON u.id=q.id
WHERE levenshtein(u.name,'Jill')<3
OR levenshtein(q.ALIAS,'Jill')<3;
#2
2
As usual, there is more than one solution:
像往常一样,有多个解决方案:
select u.id
from users u
where 3 >
any
(
select levenshtein ( 'Willey'::text, a )
from regexp_split_to_table
(
concat_ws ( ',' , u.name::text , u.alias::text )
, ','
) as a
)