Question

我在使用左外连接的报告SQL中遇到问题。

 SELECT * FROM xx.aa 
           LEFT OUTER JOIN xx.yyy
           ON aa.some_column_1 = yyy.some_column_1
           AND aa.some_column_2 = yyy.some_column_2
           AND aa.some_column_3 = yyy.some_column_3
           AND yyy.some_year = '2015'
           AND yyy.some_column_4 IS NOT NULL
           AND yyy.some_column_4 > 0
 WHERE aa.some_column_1 = '1234' 
   AND aa.some_column_2 = 'SOME_VALUE'
   AND aa.end_date >= TO_DATE('01-JAN-2015','DD-MON-YYYY')
   AND aa.start_date <= TO_DATE('31-DEC-2015','DD-MON-YYYY')
   AND (aa.some_column_3 = 'SOME_VALUE')

表aa

some_column_1 some_column_2 some_column_3 start_date end_date
1234          SOME_VALUE    SOME_VALUE    1/1/2011   6/30/2014
1234          SOME_VALUE    SOME_VALUE    7/1/2014   6/30/2015

表yyy

some_column_1 some_column_2 some_column_3 some_column_4 some_year
1234          SOME_VALUE    SOME_VALUE    444           2015

但是，左外连接中的最后两个条件IS NOT NULL和> 0应仅在超过1行匹配且仅至少其中一行满足两个条件时应用。否则这将是一个问题。

有什么想法吗？

Answer 1

您应该在左连接中没有条件的情况下运行查询，然后对您的条件进行讨论，以确定yyy表中的值是否为null。

select aa_col1, aa_col2, ..., aa_coln,
    case when cnt > 2 and cond > 1 
              and not (yyy_some_column_4 IS NOT NULL AND yyy_some_column_4 > 0) 
         then null else yyy_col1 
         end as yyy_col1,
    ,...the other columns of yyy
from(
    SELECT aa.*, yyy.*,
      count(*) over() as cnt,
      count(case when yyy.some_column_4 IS NOT NULL AND yyy.some_column_4 > 0  then 1 end) over() as cond
    FROM xx.aa 
    LEFT OUTER JOIN xx.yyy
        ON aa.some_column_1 = yyy.some_column_1
               AND aa.some_column_2 = yyy.some_column_2
               AND aa.some_column_3 = yyy.some_column_3
               AND yyy.some_year = '2015'
     WHERE aa.some_column_1 = '1234' 
       AND aa.some_column_2 = 'SOME_VALUE'
       AND aa.end_date >= TO_DATE('01-JAN-2015','DD-MON-YYYY')
       AND aa.start_date <= TO_DATE('31-DEC-2015','DD-MON-YYYY')
       AND (aa.some_column_3 = 'SOME_VALUE')
)

在第8行，您应该为列添加别名，使它们在上部选择中不同。我把它们命名为aa_col1，yyy_col1等等。

Oracle留下外连接问题

1 个答案: