Sql 在Oracle中查找数据子集的步骤
我在emp1中有一些记录:Sql 在Oracle中查找数据子集的步骤,sql,oracle,Sql,Oracle,我在emp1中有一些记录: SELECT distinct substrb(emp.employee_NAME,1,50) employee_NAME FROM employee emp , employee_sites sites , (SELECT DISTINCT employee_id , emp_site_number FROM abc ) abc where emp.employee_id = sites.employee_id and abc.employee_id=emp.e
SELECT distinct
substrb(emp.employee_NAME,1,50) employee_NAME
FROM employee emp , employee_sites sites , (SELECT DISTINCT employee_id ,
emp_site_number
FROM abc
) abc
where emp.employee_id = sites.employee_id
and abc.employee_id=emp.employee_id
and abc.emp_site_number = sites.emp_site_number ;
以及emp中的一些记录:
SELECT distinct emp.employee_NAME employee_NAME
FROM employee emp
WHERE 1=1 and EXISTS
(SELECT 1 FROM employee_ACCOUNTS acc WHERE acc.employee_id = emp.employee_id
)
emp的行数:205001
emp1的行数:18003
我想知道emp是否拥有emp1的所有记录,换句话说,emp是否是emp1的超集。我试过这个:
select count(*) from (SELECT distinct emp.employee_NAME employee_NAME
FROM employee emp
WHERE 1=1 and EXISTS
(SELECT 1 FROM employee_ACCOUNTS acc WHERE acc.employee_id = emp.employee_id
) ) emp ,
(SELECT distinct
substrb(emp.employee_NAME,1,50) employee_NAME
FROM employee emp , employee_sites sites , (SELECT DISTINCT employee_id ,
emp_site_number
FROM abc
) abc
where emp.employee_id = sites.employee_id
and abc.employee_id=emp.employee_id
and abc.emp_site_number = sites.emp_site_number) emp1
where emp.employee_NAME = emp1.employee_NAME ;
上述查询的行数:12360。
所以我得出结论,emp不是emp1的超集
有人请让我知道我所做的是好的,或者它需要一些修改。
如果你知道更好的方法,也请分享
谢谢您可以避免相关的子查询,只需执行一个简单的集减操作:
select employee_name -- or whatever makes the employee the same in 2 tables
from emp1 -- the table which may have rows not in the other table
MINUS
select employee_name
from emp2 -- the table which you think may be missing some rows
也可以使用左连接:
select emp2.employee_name from emp2
left join emp1 on emp2.employee_name = emp1.employee_name
where emp1.employee_name is null
性能将取决于索引、数据量等因素。检查查询计划和基准测试将让您很好地了解哪一个是更好的选择。您可以避免相关的子查询,只需执行一个简单的集减操作:
select employee_name -- or whatever makes the employee the same in 2 tables
from emp1 -- the table which may have rows not in the other table
MINUS
select employee_name
from emp2 -- the table which you think may be missing some rows
也可以使用左连接:
select emp2.employee_name from emp2
left join emp1 on emp2.employee_name = emp1.employee_name
where emp1.employee_name is null
性能将取决于索引、数据量等因素。检查查询计划和基准测试可以让您很好地了解哪一个是更好的选择。我不太喜欢反连接。我发现它们的可读性不如直接的
不存在
和不存在
。我想这就是为什么您在第二个查询中混淆了这两个表:-)无论如何,在我发现的所有方法中,减法是最干净的,这也是我的第一种方法。我不太喜欢反联接。我发现它们的可读性不如直接的不存在
和不存在
。我想这就是为什么您在第二个查询中混淆了这两个表:-)无论如何,在我发现的所有方法中,减法是最干净的,这也是我的第一种方法。