如何比较 2 列并返回 oracle SQL 中的差异

Question

我们在 oracle SQL 中的一张表中有 2 列 Col1="桌子上有书" Col2= "椅子上有花" 现在我需要将结果作为 column3 中的不同数据作为新列 col3。 col3 结果应该是 “都是花椅”。 如何在 oracle SQL 中实现这一点？？

Answer 1

你可以使用：

WITH words ( rid, col, name, id, word ) AS (
  SELECT rid,
         CASE INSTR(col, ' ')
         WHEN 0
         THEN NULL
         ELSE SUBSTR(col, INSTR(col, ' ') + 1)
         END,
         name,
         1,
         CASE INSTR(col, ' ')
         WHEN 0
         THEN col
         ELSE SUBSTR(col, 1, INSTR(col, ' ') - 1)
         END
  FROM   ( SELECT ROWID AS rid, col1, col2 FROM table_name )
  UNPIVOT ( col FOR name IN (col1, col2) )
UNION ALL
  SELECT rid,
         CASE INSTR(col, ' ')
         WHEN 0
         THEN NULL
         ELSE SUBSTR(col, INSTR(col, ' ') + 1)
         END,
         name,
         id + 1,
         CASE INSTR(col, ' ')
         WHEN 0
         THEN col
         ELSE SUBSTR(col, 1, INSTR(col, ' ') - 1)
         END
  FROM   words
  WHERE col IS NOT NULL
),
paired_words ( rid, id1, id2 ) AS (
  SELECT c1.rid,
         c1.id AS id1,
         c2.id AS id2
  FROM   ( SELECT rid, id, word FROM words WHERE name = 'COL1' ) c1
         INNER JOIN
         ( SELECT rid, id, word FROM words WHERE name = 'COL2' ) c2
         ON (c1.rid = c2.rid AND c1.word = c2.word)
),
max_path ( rid, path ) AS (
  SELECT rid,
         path
  FROM   (
    SELECT rid,
           SYS_CONNECT_BY_PATH(id2, ',') || ',' AS path,
           ROW_NUMBER() OVER (PARTITION BY rid ORDER BY LEVEL DESC) AS rn
    FROM   paired_words
    CONNECT BY PRIOR rid = rid
    AND    PRIOR id1 < id1
    AND    PRIOR id2 < id2
  )
  WHERE  rn = 1
)
SELECT LISTAGG(word, ' ') WITHIN GROUP (ORDER BY id) AS missing
FROM   words w
WHERE  NOT EXISTS (
         SELECT 1
         FROM   max_path mp
         WHERE  w.rid = mp.rid
         AND    mp.path LIKE '%,' || w.id || ',%'
       )
AND    w.name = 'COL2'
GROUP BY rid;

其中，对于样本数据：

CREATE TABLE table_name ( col1, col2 ) AS
SELECT 'there is book on the table', 'there are flowers on the chair' FROM DUAL UNION ALL
SELECT 'there is book on the table', 'there is a book on the table' FROM DUAL UNION ALL
SELECT 'there is book on the table', 'there is book there is book on the table on the table' FROM DUAL

输出：

MISSING
are flowers chair
a
there is book on the table

db小提琴here

Answer 2

这是一个选项（按照您的要求）。在代码中读取 cmets。

SQL> with test (id, col1, col2) as
  2    (select 1, 'there is book on the table',
  3               'there are flowers on the chair'
  4     from dual
  5    ),
  6  -- split sentences into words (each in its own line)
  7  sent1 as
  8    (select id,
  9         column_value cv,
 10         regexp_substr(col1, '[^ ]+', 1, column_value) word
 11     from test cross join
 12       table(cast(multiset(select level from dual
 13                           connect by level <= regexp_count(col1, ' ') + 1
 14                          ) as sys.odcinumberlist))
 15    ),
 16  sent2 as
 17    (select id,
 18         column_value cv,
 19         regexp_substr(col2, '[^ ]+', 1, column_value) word
 20     from test cross join
 21       table(cast(multiset(select level from dual
 22                           connect by level <= regexp_count(col2, ' ') + 1
 23                          ) as sys.odcinumberlist))
 24    )
 25  -- final result
 26  select a.id,
 27         listagg(b.word, ' ') within group (order by a.cv) result
 28  from sent2 b join sent1 a on a.id = b.id and a.cv = b.cv and a.word <> b.word
 29  group by a.id;

        ID RESULT
---------- ------------------------------
         1 are flowers chair

SQL>