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From Saurabh S <>
Subject Accessing elements from array returned by split() function
Date Thu, 01 Mar 2012 21:19:02 GMT


I have a set of URLs which I need to parse. For example, if the url is,,

I need to extract, 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.


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