选择TOP N和BOTTOM N [英] Select TOP N and BOTTOM N
问题描述
尝试获取前n行,后n行.尽管它给了我结果,但是却需要很多时间.我相信它会扫描表两次.
Trying to fetch top n bottom n rows. Though it gives me result but, it takes lot of time. I believe it scans table twice.
Code used:
WITH TI AS
(SELECT * FROM
(SELECT
Column1,
Column2,
Colmn3
FROM TABLE
ORDER BY DESC
)
WHERE ROWNUM<=5),
T2 AS
(SELECT * FROM
(SELECT
Column1,
Column2,
Colmn3
FROM TABLE
ORDER BY ASC
)
WHERE ROWNUM<=5)
SELECT * FROM T1
UNION ALL
SELECT * FROM T2
我该如何以更快的方式获取呢? 考虑到表会定期更新.
How can i fetch this in more faster way?? Considering that tables are updated regularly.
推荐答案
解决此问题的最佳方法部分取决于您的Oracle版本.这是使用版本12.1中添加的match_recognize
子句的非常简单(并且,我怀疑是非常有效的)解决方案.
The best way to solve this problem depends in part on your Oracle version. Here is a very simple (and, I suspect, very efficient) solution using the match_recognize
clause, added in version 12.1.
我使用标准HR
架构中的EMPLOYEES
表来说明它,并按SALARY
进行排序.这里唯一的技巧是选择顶部和底部的五行,而忽略它们之间的所有内容. (忽略")是{- ... -}
运算符在pattern
子句中所做的.
I illustrate it using the EMPLOYEES
table in the standard HR
schema, ordering by SALARY
. The only trick here is to select the top and bottom five rows, and to ignore everything in between; that (the "ignoring") is what the {- ... -}
operator does in the pattern
sub-clause.
select employee_id, first_name, last_name, salary
from hr.employees
match_recognize(
order by salary desc
all rows per match
pattern ( a{5} {- a* -} a{5} )
define a as 0 = 0 -- For reasons known only to Oracle, DEFINE is required.
);
EMPLOYEE_ID FIRST_NAME LAST_NAME SALARY
----------- -------------------- ------------------------- ----------
100 Steven King 24000
101 Neena Kochhar 17000
102 Lex De Haan 17000
145 John Russell 14000
146 Karen Partners 13500
135 Ki Gee 2400
127 James Landry 2400
136 Hazel Philtanker 2200
128 Steven Markle 2200
132 TJ Olson 2100
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