与 Oracle 的 in join on 子句没有区别

Question

我有一个使用左外连接构造的查询，如下所示：

left outer JOIN GA_LOAN GA 
  ON LOAN.LOAN_TYPE = GA.LOAN_TYP
    AND LOAN.DT = GA.GUARANTY_DT
    AND LOAN.FFEL_DUP_ID = GA.SEP_LOAN_IND
    AND LOAN.SCH_BR_CODE = GA.ORIG_SCHL_CD
    AND STU.CURR_SSN = GA.STU_SSN
    AND STU.DOB = GA.DOB
    and stu.curr_fst = ga.stu_first_nam
  --and (plus_bor.curr_ssn is not distinct from ga.plus_brwr_ssn )

当我添加注释掉的行时，出现以下错误。

ORA-00908: missing NULL keyword
00908. 00000 -  "missing NULL keyword"
*Cause:    
*Action:

与 DB2 中这种结构的正常工作没有什么不同，但 Oracle 给我带来了问题。有什么建议么？

is not distinct from如果我用 a替换，我不会收到任何错误=，但这在逻辑上是不一样的。

is not distinct from如果两个值都是 ，则 with 给出匹配项null，而 as=在这种情况下不匹配。

Answer 1

IS DISTINCT FROM您可以通过NOT EXISTS结合使用来模拟INTERSECT：

plus_bor.curr_ssn IS DISTINCT FROM ga.plus_brwr_ssn
<=>
NOT EXISTS (SELECT plus_bor.curr_ssn FROM dual INTERSECT
            SELECT ga.plus_brwr_ssn FROM dual);

例子：

WITH cte(a,b) AS (
    SELECT 1, NULL    FROM dual UNION ALL
    SELECT 1,2        FROM dual UNION ALL
    SELECT 1,1        FROM dual UNION ALL
    SELECT NULL, 1    FROM dual UNION ALL
    SELECT NULL, NULL FROM dual
)
SELECT *
FROM cte
WHERE NOT EXISTS (SELECT a FROM dual INTERSECT
                SELECT b FROM dual)

Rextester 演示

输出：

A       B
------------
1       NULL
1       2
NULL    1

在你的情况下IS NOT DISTINCT FROM很简单EXISTS：

plus_bor.curr_ssn IS NOT DISTINCT FROM ga.plus_brwr_ssn
<=>
EXISTS (SELECT plus_bor.curr_ssn FROM dual INTERSECT
        SELECT ga.plus_brwr_ssn FROM dual);

例子：

WITH cte(a,b) AS (
    SELECT 1, NULL      FROM dual UNION ALL
    SELECT 1,2          FROM dual UNION ALL
    SELECT 1,1          FROM dual UNION ALL
    SELECT NULL, 1      FROM dual UNION ALL
    SELECT NULL, NULL   FROM dual
)
SELECT *
FROM cte
WHERE EXISTS (SELECT a FROM dual INTERSECT
              SELECT b FROM dual);

输出：

 A      B
 1      1
 NULL   NULL

雷克斯特演示2

附录

与评论中提出的方法相比，这种方法有一个很大的优势COALESCE/NVL。您不必考虑依赖于数据类型的默认中性值。

例如，如果列是数据类型//DATE那么你必须编写如下内容：INTTEXT

coalesce(col1,DATE '1900-01-01') = coalesce(col2,DATE '1900-01-01')
coalesce(col1, 0) = coalesce(col2, 0)
coalesce(col1, ' ') = coalesce(col2, ' ')

当然，碰撞的可能性很小。例如：

coalesce(col1, 0) = coalesce(col2, 0)
=>
col1 = NULL
col2 = 0
and we have incorrect match!!!