tạo Oracle View để so sánh dữ liệu dựa trên các điều kiện

Bạn có thể kiểm tra truy vấn sau không, tôi đã thực hiện tất cả các trường hợp trong with và sau đó kết hợp ra khỏi nó.

Chỉnh sửa:- Sau khi thảo luận và làm rõ với OP qua trò chuyện

Chúng tôi không cần full join bất kỳ nữa và bằng cách truy cập bảng theo từng trường hợp, nó được viết lại.

-- case 1
-- when fund_isin with member_ratio = 'O' present in both is_id and is_id_tst table
-- and the value of is_id.member_ratio < is_id_tst.memebr_ratio
-- logic --
-- the from clasuse says take all the records from is_id table 
-- by corelate the fund_isin (t1.fund_isin = t.fund_isin)
-- the subquery then finds record by joining both table is_id and is_id_tst for member_ratio = 'O'
-- and where the member_ratio is smaller (is_id_tst.member_ratio > is_id.member_ratio)
-- extra condition on is_id_tst table is the member_ratio value should be greater than 0 for member_descr='O'
WITH ratio_lower_is_id
AS
(SELECT *
   FROM is_id t
  WHERE EXISTS 
  (SELECT 1
     FROM is_id_tst t2
     JOIN is_id t1
       ON t2.fund_isin = t1.fund_isin
    WHERE t1.fund_isin = t.fund_isin
      AND t2.member_descr = 'O'
      AND t1.member_descr = 'O'
      AND t2.member_ratio > 0
      AND t2.member_ratio > 
          t1.member_ratio)
),
-- case 2
-- applies the same logic as in case 1 but then take records from is_id_tst table
-- where the member_ratio having lower value for record with member_descr='O'
-- in comparison with the record present in is_id table for memebr_descr='O'
ratio_lower_is_id_tst
AS
(SELECT *
   FROM is_id_tst t
  WHERE t.member_ratio > 0
    AND EXISTS 
  (SELECT 1
     FROM is_id t2
     JOIN is_id_tst t1
       ON t2.fund_isin = t1.fund_isin
    WHERE t1.fund_isin = t.fund_isin
      AND t2.member_descr = 'O'
      AND t1.member_descr = 'O'
      AND t2.member_ratio > 
          t1.member_ratio)
),
-- case 3
-- take all records from is_id_tst table for all each unique fund_isin 
-- where the member_ratio value is < 0 for record member_descr='O'
-- and is avaialble in is_id_tst table irrespective of what record for the same
-- fund_isin available in is_id table
ratio_minus_is_id_tst
AS
(SELECT *
   FROM is_id_tst t
  WHERE EXISTS 
  (SELECT 1
     FROM is_id_tst t1
    WHERE t1.fund_isin = t.fund_isin
      AND t1.member_descr = 'O'
      AND t1.member_ratio < 0)
),
-- case 4
-- take all the records from is_id table 
-- where the fund_isin is not available in is_id_tst table
only_in_is_id
AS
(
SELECT *
  FROM is_id t1
 WHERE NOT EXISTS
   (SELECT 1 
      FROM is_id_tst t2
     WHERE t2.fund_isin = t1.fund_isin)
),
-- case 5
-- take all the records from is_id_tst table
-- where the fund_isin is not available in is_id table
only_in_is_id_tst
AS
(
SELECT *
  FROM is_id_tst t1
 WHERE NOT EXISTS
   (SELECT 1 
      FROM is_id t2
     WHERE t2.fund_isin = t1.fund_isin)
)
-- finally once all the sets as per each case available
-- take each of them and do a union all for the final result set
-- one level sub query required only if we want to sort the result otherwise can be removed
-- and only union all of all sets from with clause is enough
select *
  from 
(
-- case1
select *
  from ratio_lower_is_id
union all
-- case 2
select *
  from ratio_lower_is_id_tst
union all
-- case 3
select *
  from ratio_minus_is_id_tst
union all
-- case 4
select *
  from only_in_is_id
union all
-- case 5
select *
  from only_in_is_id_tst
)
order by fund_isin;