oracle中按部分字符串分组数据sql（Oracle9i企业版Release 9.2.0.4.0）

Question

我有以下查询：

select referrer, count(distinct ad_id) as Adverts,
       sum(case f when 'Y' then hits else 0 end) as clicks,
       sum(case f when 'N' then hits else 0 end) as views
from advert_view_hits
where ad_id in ({$id_strings})
group by referrer"

这将 return 如下数据：

 Referrer  Adverts  Clicks  Views
 Caterer     3       124     74
 Indeed      5       234     136

这很好，但在某些情况下，引荐来源网址已像这样存储在数据库中：

user1@jwrecruitment.co.uk_200890,
user2@jwrecruitment.co.uk_200890

user1@gatewayjobs.co.uk_200890, 
user3@towngate-personnel.co.uk_2

我将如何根据用作推荐人的用户电子邮件公司对数据进行分组。

因此数据将如下所示：

Referrer             Adverts  Clicks  Views
 Caterer               3       124     74
 Indeed                5       234     136
 jwrecruitment.co.uk   8       456     782 
 gatewayjobs.co.uk     9       897     959

这样像 jwrecruitment.co.uk 这样的电子邮件的所有数据将被组合在一起并显示。

Answer 1

如果我没听错，你可以使用 regexp_replace() :

select 
    regexp_replace(referrer, '^.*@([^_]+).*$', '') referrer, 
    count(distinct ad_id) as Adverts,
    sum(case f when 'Y' then hits else 0 end) as clicks,
    sum(case f when 'N' then hits else 0 end) as views
from advert_view_hits
where ad_id in ({$id_strings})
group by regexp_replace(referrer, '^.*@([^_]+).*$', '')

正则表达式匹配包含 arobas 的引荐来源网址，并捕获 arobas 之后下划线之前的部分。如果 referrer 与正则表达式不匹配，则保持不变。

Answer 2

如果我没理解错的话，你可以在 @ 之后获取所有内容——你可以使用 regexp_substr():

select regexp_substr(referrer, '[^@]+$') as referrer, count(distinct ad_id) as Adverts,
       sum(case f when 'Y' then hits else 0 end) as clicks,
       sum(case f when 'N' then hits else 0 end) as views
from advert_view_hits
where ad_id in ({$id_strings})
group by regexp_substr(referrer, '[^@]+$') ;

您可以将 regexp_substr() 逻辑替换为：

select substr(referrer, instr(referrer, '@') + 1) as referrer, count(distinct ad_id) as Adverts,
       sum(case f when 'Y' then hits else 0 end) as clicks,
       sum(case f when 'N' then hits else 0 end) as views
from advert_view_hits
where ad_id in ({$id_strings})
group by  substr(referrer, instr(referrer, '@') + 1) ;

Answer 3

您也可以使用regexp_substr来查找@和_之间的字符串，如下所示：

select REGEXP_SUBSTR(referrer,'@([^_]+)',1,1,NULL,1) referrer, 
       count(distinct ad_id) as Adverts,
       sum(case f when 'Y' then hits else 0 end) as clicks,
       sum(case f when 'N' then hits else 0 end) as views
  from advert_view_hits
where ad_id in ({$id_strings})
group by REGEXP_SUBSTR(referrer,'@([^_]+)',1,1,NULL,1)

如果您使用的是旧版本，则不要使用 regexp_substr，而是使用 SUBSTR 和 INSTR 的组合，如下所示：

SUBSTR(referrer, 
       INSTR(referrer, '@') + 1, 
       DECODE(INSTR(referrer, '_', - 1), 
              0, 
              LENGTH(referrer) - INSTR(referrer, '@'), 
              INSTR(referrer, '_', - 1) - INSTR(referrer, '@') - 1)
      )

oracle中按部分字符串分组数据sql（Oracle9i企业版Release 9.2.0.4.0）

Group data by part of string in oracle sql ( Oracle9i Enterprise Edition Release 9.2.0.4.0)

sql

oracle

group-by

count

oracle9i