Sql server 2005 SQLServer2005中的分析函数
我有一个查询,它在ORACLE中工作,但在SQL SERVER 2005中不工作…如何将此查询转换为在SQL SERVER 2005中工作Sql server 2005 SQLServer2005中的分析函数,sql-server-2005,oracle,Sql Server 2005,Oracle,我有一个查询,它在ORACLE中工作,但在SQL SERVER 2005中不工作…如何将此查询转换为在SQL SERVER 2005中工作 select user_name , url , count(*) ,first_value(count(*)) over (partition by user_name order by count(*) desc) max_total fr
select user_name
, url
, count(*)
,first_value(count(*)) over (partition by user_name
order by count(*) desc) max_total
from urls
group by user_name, url
order by max_total desc,user_name
结果:
由于我的评论很大,我将把它作为答案写下来:( @Tang虽然不太正确,但有一条正确的路径…也许我已经通过查询解决了这个问题,但我希望通过分析函数解决它…我的查询是这样的
select urls.user_name
,urls.url
,count(*) ct
,max_amount
from urls
,(select user_name
,max(amount) max_amount
from (select user_name
,url
,count(*) amount
from urls
group by user_name,url) t1
group by user_name) t2
where urls.user_name=t2.user_name
group by urls.user_name,urls.url,max_amount
order by max_amount desc,urls.user_name,ct desc
@这是测试数据
create table urls(
user_name varchar2(100),
url varchar2(100)
);
insert into urls
values('mariami','google.com');
insert into urls
values('mariami','google.com');
insert into urls
values('mariami','google.com');
insert into urls
values('giorgi','google.com');
insert into urls
values('giorgi','google.com');
insert into urls
values('giorgi','facebook.com');
insert into urls
values('giorgi','facebook.com');
insert into urls
values('giorgi','facebook.com');
insert into urls
values('giorgi','facebook.com');
insert into urls
values('mariami','facebook.com');
insert into urls
values('a','facebook.com');
我的查询结果是:
您的查询结果是:
您能描述一下查询应该返回什么吗?我已经上传了文件,其中显示了结果。您的查询和我的查询都是相同的,您有第二个子查询,我通过MAX解析。如果您运行您的查询和我的查询,结果是相同的。不,不,它不一样(您应该计算每个用户名的MAX)我会给你测试数据看看区别…测试数据是向上的…!瞧!你是对的,我选择了退出第二个内部查询,而事实上它是necessary@tanging谢谢你和我分享的时间我真的很感谢你的帮助谢谢你,你检查过这个了吗?很遗憾,我现在无法检查您的答案,我没有sql server。但明天我将能够检查。。。
WITH q AS
(
SELECT user_name, url, COUNT(*) AS cnt
FROM urls
GROUP BY
user_name, url
)
SELECT *
FROM q qo
CROSS APPLY
(
SELECT TOP 1 cnt
FROM q qi
WHERE qi.user_name = qo.user_name
ORDER BY
cnt DESC
) qi