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 >
