oracle sql 通過打斷字符串並比較每個單詞來匹配兩列中的單詞

Question

我的表中有兩列，例如column_A和column_B

假設column_A有數據“best buy card credit”， column_B有“credit no take buy order”，這里我需要匹配兩列中的單詞並返回匹配單詞的數量。 在這種情況下，它應該返回 2 作為“購買”和“信用”匹配。 任何人都可以建議 sql 代碼來做同樣的事情。

請注意： column_a和column_b大小不是固定的，即兩者中的字數可能會改變。

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