有没有办法使一个Oracle查询行为像它包含一个MySQL限制子句?
在MySQL中，我可以这样做:
select *
from sometable
order by name
limit 20,10
要得到第21行到第30行(跳过前20行，给出接下来的10行)。这些行是按顺序选择的，所以实际上是从第20个名字的字母顺序开始的。
在Oracle中，人们唯一提到的是rownum伪列，但它在order by之前求值，这意味着:
select *
from sometable
where rownum <= 10
order by name
将返回一个随机的10行按名称排序的集合，这通常不是我想要的。它也不允许指定偏移量。
当前回答
我为以下方法做了一些性能测试:
Asktom
select * from (
select a.*, ROWNUM rnum from (
<select statement with order by clause>
) a where rownum <= MAX_ROW
) where rnum >= MIN_ROW
分析
select * from (
<select statement with order by clause>
) where myrow between MIN_ROW and MAX_ROW
短的替代
select * from (
select statement, rownum as RN with order by clause
) where a.rn >= MIN_ROW and a.rn <= MAX_ROW
结果
表有1000万条记录，排序在一个未索引的datetime行上:
解释计划对所有三个选择显示相同的值(323168) 但是赢家是AskTom (analytic紧随其后)
选择前10行需要:
AskTom: 28-30秒 分析型:33-37秒 短选择:110-140秒
在100,000和100,010之间选择行:
AskTom: 60秒 分析性:100秒
在9,000,000和9,000,010之间选择行:
AskTom: 130秒 分析型:150秒
其他回答
在Oracle 12c上(参见SQL参考中的行限制子句):
SELECT *
FROM sometable
ORDER BY name
OFFSET 20 ROWS FETCH NEXT 10 ROWS ONLY;
您可以为此使用子查询
select *
from
( select *
from emp
order by sal desc )
where ROWNUM <= 5;
还可以查看Oracle/AskTom上的关于ROWNUM和限制结果的主题以获得更多信息。
更新: 为了同时限制结果的下界和上界，事情变得有点臃肿
select * from
( select a.*, ROWNUM rnum from
( <your_query_goes_here, with order by> ) a
where ROWNUM <= :MAX_ROW_TO_FETCH )
where rnum >= :MIN_ROW_TO_FETCH;
(复制自指定的AskTom-article)
更新2: 从Oracle 12c(12.1)开始，就有了限制行或从偏移量开始的语法。
SELECT *
FROM sometable
ORDER BY name
OFFSET 20 ROWS FETCH NEXT 10 ROWS ONLY;
更多的例子请看这个答案。谢谢克鲁米亚的提示。
我已经开始准备Oracle 1z0-047考试，与12c进行了验证 在准备它的时候，我遇到了一个被称为“FETCH FIRST”的12c增强。 它允许您根据您的方便获取行/限制行。 它有几个可用的选项
- FETCH FIRST n ROWS ONLY
- OFFSET n ROWS FETCH NEXT N1 ROWS ONLY // leave the n rows and display next N1 rows
- n % rows via FETCH FIRST N PERCENT ROWS ONLY
例子:
Select * from XYZ a
order by a.pqr
FETCH FIRST 10 ROWS ONLY
SQL标准
从版本12c开始，Oracle支持SQL:2008标准，该标准提供了以下语法来限制SQL结果集:
SELECT
title
FROM
post
ORDER BY
id DESC
FETCH FIRST 50 ROWS ONLY
Oracle 11g及更老版本
在版本12c之前，要获取Top-N记录，必须使用派生表和ROWNUM伪列:
SELECT *
FROM (
SELECT
title
FROM
post
ORDER BY
id DESC
)
WHERE ROWNUM <= 50
作为公认答案的扩展，Oracle内部使用ROW_NUMBER/RANK函数。OFFSET FETCH语法是语法糖。
可以使用DBMS_UTILITY来观察。EXPAND_SQL_TEXT过程:
准备样例:
CREATE TABLE rownum_order_test (
val NUMBER
);
INSERT ALL
INTO rownum_order_test
SELECT level
FROM dual
CONNECT BY level <= 10;
COMMIT;
查询:
SELECT val
FROM rownum_order_test
ORDER BY val DESC
FETCH FIRST 5 ROWS ONLY;
常规:
SELECT "A1"."VAL" "VAL"
FROM (SELECT "A2"."VAL" "VAL","A2"."VAL" "rowlimit_$_0",
ROW_NUMBER() OVER ( ORDER BY "A2"."VAL" DESC ) "rowlimit_$$_rownumber"
FROM "ROWNUM_ORDER_TEST" "A2") "A1"
WHERE "A1"."rowlimit_$$_rownumber"<=5 ORDER BY "A1"."rowlimit_$_0" DESC;
db < > fiddle演示
获取展开的SQL文本:
declare
x VARCHAR2(1000);
begin
dbms_utility.expand_sql_text(
input_sql_text => '
SELECT val
FROM rownum_order_test
ORDER BY val DESC
FETCH FIRST 5 ROWS ONLY',
output_sql_text => x);
dbms_output.put_line(x);
end;
/
WITH TIES扩展为RANK:
declare
x VARCHAR2(1000);
begin
dbms_utility.expand_sql_text(
input_sql_text => '
SELECT val
FROM rownum_order_test
ORDER BY val DESC
FETCH FIRST 5 ROWS WITH TIES',
output_sql_text => x);
dbms_output.put_line(x);
end;
/
SELECT "A1"."VAL" "VAL"
FROM (SELECT "A2"."VAL" "VAL","A2"."VAL" "rowlimit_$_0",
RANK() OVER ( ORDER BY "A2"."VAL" DESC ) "rowlimit_$$_rank"
FROM "ROWNUM_ORDER_TEST" "A2") "A1"
WHERE "A1"."rowlimit_$$_rank"<=5 ORDER BY "A1"."rowlimit_$_0" DESC
和偏移量:
declare
x VARCHAR2(1000);
begin
dbms_utility.expand_sql_text(
input_sql_text => '
SELECT val
FROM rownum_order_test
ORDER BY val
OFFSET 4 ROWS FETCH NEXT 4 ROWS ONLY',
output_sql_text => x);
dbms_output.put_line(x);
end;
/
SELECT "A1"."VAL" "VAL"
FROM (SELECT "A2"."VAL" "VAL","A2"."VAL" "rowlimit_$_0",
ROW_NUMBER() OVER ( ORDER BY "A2"."VAL") "rowlimit_$$_rownumber"
FROM "ROWNUM_ORDER_TEST" "A2") "A1"
WHERE "A1"."rowlimit_$$_rownumber"<=CASE WHEN (4>=0) THEN FLOOR(TO_NUMBER(4))
ELSE 0 END +4 AND "A1"."rowlimit_$$_rownumber">4
ORDER BY "A1"."rowlimit_$_0"
