You can also use parse_url(url, 'HOST') instead of a regular expression. On Thu, Mar 1, 2012 at 1:32 PM, Saurabh S <saurab...@live.com> wrote:
> Of course, it works fine now. I feel like an idiot. > > And that problem using parse_url also went away and I can use that as well. > > Thanks a bunch, Phil. > > > Date: Thu, 1 Mar 2012 21:22:27 +0000 > > Subject: Re: Accessing elements from array returned by split() function > > From: philip.j.trom...@gmail.com > > To: user@hive.apache.org > > > > > I guess that split(...)[1] is giving you what's inbetween the 1st and > > 2nd '/' character, which is nothing. Try split(...)[2]. > > > > Phil. > > > > On 1 March 2012 21:19, Saurabh S <saurab...@live.com> wrote: > > > Hello, > > > > > > I have a set of URLs which I need to parse. For example, if the url is, > > > http://www.google.com/anything/goes/here, > > > > > > I need to extract www.google.com, i.e. everything between second and > third > > > forward slashes. > > > > > > I can't figure out the regex pattern to do so, and am trying to use > split() > > > function instead. So, my hive query looks like > > > select url, split(url,'/') > > > ... > > > > > > The second column contains the entire array returned by the split > function. > > > Is there any way to access only the second element of the array, which > will > > > give me what I need? > > > > > > When I try the following statement select url, split(url,'/')[1], I > get an > > > empty second column. > > > > > > Is this the expected behavior? Any other suggestions on how to parse > the > > > URL? > > > > > > Oh by the way, I'm aware that the function parse_url(url,'HOST') will > give > > > me something similar to what I want, but for some reason, that > function on > > > my database is running extremely slow. > > > > > > First time posting to this list. If there is anything wrong, please > let me > > > know. > > > > > > Regards, > > > Saurabh > > > >
