oracle sql 通过打断字符串并比较每个单词来匹配两列中的单词

Question

I have two columns in my table for instance column_A and column_B

Suppose column_A has data "best buy card credit" and column_B has "credit no take buy order", here i need to match the words in both the columns and return the number of matching words . In this case it should return 2 as "buy" and "credit" match. Can anyone please suggest sql code to do the same.

Please NOTE : size of column_a and column_b is not fixed ie the number of words in both might change.

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