从MySQL查询在Hive中实现不平等联接

问题描述

我正在尝试用MysqL编写的蜂巢实现查询。我知道配置单元不支持ON条件下的不平等联接。下面是我的代码，并告诉我一种实现方法。

Select test1.a,test2.b,test4.c,dummy.c
from
test1 join test2 on test1.id = test2.id and test2 != 'ABC'
join test3 on test1.id = test2.id and test3 != 'Archive'
join test4 on test3.id = test4.id and test4 = 'XYZ'
left outer join
(select test1.a,test2,b
test3.c
from test1 join test2 on test1.id = test2.id and test2 != 'ABC'
join test3 on test1.id = test2.id) dummy
on test3.id = dummy.id
**and (test4.id != 1001 or dummy.c = TRUE)**
left join test5 on test3.id= test5.id
and dummy.c = TRUE

现在，用*突出显示的条件是我需要知道如何在配置单元中实现的部分，因为我无法在ON条件下实现它，并且如果将其放在where子句结果不匹配的地方。 任何建议将其重写为蜂巢将不胜感激。

解决方法

对于从LEFT JOIN中选择的列，我在SELECT语句中将不等式条件用作case语句。 下面是代码-

Select test1.a,test2.b,test4.c,case when (test4.id != 1001 or nvl(dummy.c,False))= TRUE then dummy.c end as c0
from
test1 join test2 on test1.id = test2.id and test2 != 'ABC'
join test3 on test1.id = test2.id and test3 != 'Archive'
join test4 on test3.id = test4.id and test4 = 'XYZ'
left outer join
(select test1.a,test2,b
test3.c
from test1 join test2 on test1.id = test2.id and test2 != 'ABC'
join test3 on test1.id = test2.id) dummy
on test3.id = dummy.id
left join test5 on test3.id= test5.id
and dummy.c = TRUE