在POSTGRES

Question

select regexp_replace('https://www.facebook.com/cricket/hello', '.*\..*?\/', '')

上面的代码给了我

  

您好

而不是

  

蟋蟀/你好

我检查了Regexp检查网站并且模式是正确的。 我不知道我哪里出错了。

  

DBMS：“PostgreSQL 8.2.15（Greenplum Database 4.2.8.3 build 1）on   x86_64-unknown-linux-gnu，由GCC gcc（GCC）4.4.2汇编编译而成   2014年11月2日01:33:14“

Answer 1

试试这个：

select regexp_replace('https://www.facebook.com/cricket/hello', '.*\.[a-z]+\/', '')

也可以使用cctld：

select regexp_replace('https://www.google.co.uk/cricket/hello', '.*\.[a-z]+\/', '')

Answer 2

我假设你想要一个URL的路径部分。

我没有我的pg，但我会非常清楚地知道网址的每一部分 -

'[^:]+:\/\/[A-Za-z][-a-zA-Z0-9]*(\.[A-Za-z][-a-zA-Z0-9]*)*/'

测试：

select testval, regexp_replace ( testval,  '[^:]+:\/\/[A-Za-z][-a-zA-Z0-9]*(\.[A-Za-z][-a-zA-Z0-9]*)*/',  '')
from (
    select 'https://www.facebook.com/cricket/hello' as testval
  union all
  select 'http://a.b.co.uk/cric.ke.t/hello' as testval
  union all
  select 'ftp://a.b.com.d.e.f/relroot/cricket/hello' as testval  union all
  select 'http://www.google.co.uk/cricket/hello' as testval  
  union all
  select 'http://a.b.co.uk/cricket/hello/this/is/a/little/longer?and&it=has&args' as testval
) vals

请参阅http://sqlfiddle.com/#!15/9eecb/857/0

Answer 3

我不知道如何，但这有效

.*?\.[a-z]+\/

以Andrew Wolfe对最奇怪的URL进行查询。

select testval, regexp_replace ( testval,  '.*?\.[a-z]+\/',  '')
from (
    select 'https://www.facebook.com/cricket/hello' as testval
  union all
  select 'http://a.b.co.uk/cric.ke.t/hello' as testval
  union all
  select 'ftp://a.b.com.d.e.f/relroot/cricket/hello' as testval  union all
  select 'http://www.google.co.uk/cricket/hello' as testval  
  union all
  select 'http://a.b.co.uk/cricket/hello/this/is/a/little/longer?and&it=has&args' as testval
) vals