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SQL 查询比较两个 varray()

转载 作者:行者123 更新时间:2023-12-04 08:32:31 29 4
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我有一张员工表。其中一列是一个 varray() ,其中包含他们办公室的多个房间#。我正在寻找一个简单的查询,该查询将比较每个员工以查看他们是否共用一个办公室。

SELECT  E1.Name, E2.Name
FROM Employee E1
JOIN Employee E2
ON E1.Room = E2.Room;
这样的事情不起作用,因为 Room 列是一个 varray。我只需要第一个变量中的一个值与第二个变量中的另一个值匹配。有没有简单的方法来做到这一点?

最佳答案

假设您指的是 Oracle,您选择的查询可以是

select
E1.name as employee_1, E2.name as employee_2,
R1.column_value as the_matching_room
from employee E1
cross join table(E1.rooms) R1
join employee E2
on E2.emp_id > E1.emp_id
join table(E2.rooms) R2
on R2.column_value = R1.column_value
;
或(更有效)
with rooms_unnested$ as (
select E.emp_id, E.name, R.column_value as room
from employee E
cross join table(E.rooms) R
)
select
E1.name as employee_1, E2.name as employee_2,
E1.room as the_matching_room
from rooms_unnested$ E1
join rooms_unnested$ E2
on E2.emp_id > E1.emp_id
and E2.room = E1.room
;
这个有一个潜在的问题,即首先在员工表之间进行笛卡尔运算,然后再取消集合的嵌套:
-----------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost | Time |
-----------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1334324 | 5142484696 | 447202 | 00:00:18 |
| 1 | NESTED LOOPS | | 1334324 | 5142484696 | 447202 | 00:00:18 |
| 2 | NESTED LOOPS | | 16336 | 62926272 | 63 | 00:00:01 |
| 3 | NESTED LOOPS | | 2 | 7700 | 7 | 00:00:01 |
| 4 | TABLE ACCESS FULL | EMPLOYEE | 2 | 3850 | 3 | 00:00:01 |
| * 5 | TABLE ACCESS FULL | EMPLOYEE | 1 | 1925 | 2 | 00:00:01 |
| 6 | COLLECTION ITERATOR PICKLER FETCH | | 8168 | 16336 | 28 | 00:00:01 |
| * 7 | COLLECTION ITERATOR PICKLER FETCH | | 82 | 164 | 27 | 00:00:01 |
-----------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
------------------------------------------
* 5 - filter("E2"."EMP_ID">"E1"."EMP_ID")
* 7 - filter(VALUE(KOKBF$)=VALUE(KOKBF$))
假设您的“房间”变量可能包含重复项,还有一项调整要做——让每个员工的房间都不同,这导致我们(希望)最终查询......
with rooms_unnested$ as (
select distinct
E.emp_id, E.name, R.column_value as room
from employee E
cross join table(E.rooms) R
)
select
E1.name as employee_1, E2.name as employee_2,
E1.room as the_matching_room
from rooms_unnested$ E1
join rooms_unnested$ E2
on E2.emp_id > E1.emp_id
and E2.room = E1.room
;
...这也恰好解决了笛卡尔函数的“问题”,方法是首先(并且只有一次!)取消嵌套“房间”变量数组,然后进行 equi-hash-join:
---------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost | Time |
---------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 120 | 65 | 00:00:01 |
| 1 | TEMP TABLE TRANSFORMATION | | | | | |
| 2 | LOAD AS SELECT (CURSOR DURATION MEMORY) | SYS_TEMP_0FD9D6699_11FF28DD | | | | |
| 3 | HASH UNIQUE | | 3 | 36 | 61 | 00:00:01 |
| 4 | NESTED LOOPS | | 16336 | 196032 | 59 | 00:00:01 |
| 5 | TABLE ACCESS FULL | EMPLOYEE | 2 | 20 | 3 | 00:00:01 |
| 6 | COLLECTION ITERATOR PICKLER FETCH | | 8168 | 16336 | 28 | 00:00:01 |
| * 7 | HASH JOIN | | 1 | 120 | 4 | 00:00:01 |
| 8 | VIEW | | 3 | 180 | 2 | 00:00:01 |
| 9 | TABLE ACCESS FULL | SYS_TEMP_0FD9D6699_11FF28DD | 3 | 36 | 2 | 00:00:01 |
| 10 | VIEW | | 3 | 180 | 2 | 00:00:01 |
| 11 | TABLE ACCESS FULL | SYS_TEMP_0FD9D6699_11FF28DD | 3 | 36 | 2 | 00:00:01 |
---------------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
------------------------------------------
* 7 - access("E2"."ROOM"="E1"."ROOM")
* 7 - filter("E2"."EMP_ID">"E1"."EMP_ID")

关于SQL 查询比较两个 varray(),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64959531/

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