sql - 通过打破字符串并比较每个单词来匹配两列oracle sql中的单词

Question

column_A例如，我的表中有两列column_B

假设column_A有数据“best buy card credit”和column_B“credit no take buy order”，这里我需要匹配两列中的单词并返回匹配单词的数量。在这种情况下，它应该返回 2 作为“购买”和“信用”匹配。任何人都可以请建议sql代码来做同样的事情。

请注意：column_a和的大小column_b不是固定的，即两者中的字数可能会发生变化。

Answer 1

with t as (
SELECT 1 AS ID, 'best buy card credit' column_1,
       'credit no take buy order' column_2
  FROM DUAL
UNION ALL
SELECT 2 AS ID, 'gaurav is  fool' column_1, 'saurabh is fool' column_2
  FROM DUAL
           )
,t1_column1 as (
SELECT     ID, LEVEL AS n, REGEXP_SUBSTR (column_1, '[^ ]+', 1, LEVEL) AS val
      FROM t
CONNECT BY REGEXP_SUBSTR (column_1, '[^ ]+', 1, LEVEL) IS NOT NULL
        )
,t1_column2 as (
SELECT     ID, LEVEL AS n, REGEXP_SUBSTR (column_2, '[^ ]+', 1, LEVEL) AS val
      FROM t
CONNECT BY REGEXP_SUBSTR (column_2, '[^ ]+', 1, LEVEL) IS NOT NULL
        )

select id, LISTAGG(VAL,',') WITHIN GROUP(ORDER BY VAL ) words ,COUNT(*) "total matched words"
from
(
SELECT DISTINCT t1_column1.ID ID, t1_column1.val val
           FROM t1_column1, t1_column2
          WHERE t1_column1.ID = t1_column2.ID
            AND t1_column1.val = t1_column2.val
)
group by id