Question

我想找到减号查询的两边每个所有者的表名之间的差异。

说我从一个表中有2条select语句，我想这样比较它们：

Select1：

SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha';

Select2：

SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie';

说alpha的表有表名：a，b，c，而charlie的表有表名b，c。

如果我愿意

(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha') minus    
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie';

这给了我预期的结果：a。

相反，如果我做相反的事情：

  (SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie') minus    
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha';

这给了我预期的结果：空行。

现在，如果我想要表名“ a”和空行，我都将执行：

(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha') minus 
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie'
  Union
(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie') minus 
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha';

那我什么也没得到。

如何返回“ a”行和空行的总和？我正在使用Oracle DB

Answer 1

仅在合并后使用括号

(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha') minus 
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie'

  Union
(
(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie') minus 
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha'
)

示例

with t1 as
(
select 62054 as  sample_id,'2018-09-25 10:18:15' as timestamp,2652 as p_id from dual
union all
select 62054 as  sample_id,'2018-09-27 16:44:57' as timestamp,966 as p_id from dual
union all
select 62046 as  sample_id,null ,1809 as p_id from dual
union all
select 62046 , '018-09-25 10:18:15' ,2097  from dual



) select * from t1 where p_id=2652
 minus
 select * from t1 where p_id=966
 union

 select * from t1 where p_id=2652
 minus
 select * from t1 where p_id=2652

以上将返回null 但是

with t1 as
(
select 62054 as  sample_id,'2018-09-25 10:18:15' as timestamp,2652 as p_id from dual
union all
select 62054 as  sample_id,'2018-09-27 16:44:57' as timestamp,966 as p_id from dual
union all
select 62046 as  sample_id,null ,1809 as p_id from dual
union all
select 62046 , '018-09-25 10:18:15' ,2097  from dual



) select * from t1 where p_id=2652
 minus
 select * from t1 where p_id=966
 union
(
 select * from t1 where p_id=2652
 minus
 select * from t1 where p_id=2652
 )

上面将返回下面

SAMPLE_ID   TIMESTAMP   P_ID
62054   2018-09-25 10:18:15 2652

Answer 2

如果您明白我的意思，似乎您想要的是UNION MINUS INTERSECT。计算并集，计算相交，然后在这两个结果集之间进行MINUS（使用适当的括号，这是您当前的工作失败的原因，因为UNION和MINUS具有相同的优先级）

(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha' UNION 
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie')
MINUS
(SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'charlie' INTERSECT
SELECT TABLE_NAME FROM ALL_TABLES WHERE OWNER = 'alpha');

我相信，这正确反映了您逻辑的要求。 Zaynul Abadin Tuhin's answer解决了当前的问题。但是我相信您的努力（及其解决方案）掩盖了这里的意图。

选择减号运算符的两侧

2 个答案: