regexp_extract 在 Hive 中查找值

Question

I'm new to regexp_extract and need to split the column on / and then pick the 3rd value. For example, from

application/motorola products/routers 

would want to get routers . If there is no 3rd value, then we need to fall back on the 2nd value, which would be motorola products . I tried the following regex pattern but it doesn't work:

(.*?\/)(.*?\/)(.*?)(\/.*\/)

Answer 1

You are saying a single character is optional. Give the . a quantifier * or + . I think this regex would actually be better:

(?:([^\/]+?\/)([^\/]+?)\/([^\/]*)|([^\/]+?\/)([^\/]+))

Demo: https://regex101.com/r/dX6uQ9/2

I haven't worked with/don't have hive so can't confirm this will work but I think it should put you in a closer direction.

Answer 2

It sounds like you just want the last value, meaning whatever is after the last / . The regex for that would be [^/]+$ :

select regexp_extract(name, '[^/]+$', 0) from dummy;

If there are two slashes, you get the third value. If there are five slashes, you get the sixth value.

If you want to stop at the third value even if there are more than two slashes, you can use this:

select regexp_extract(name, '^(?:[^/]+/){0,2}([^/]+)', 1) from dummy;

The index argument, 1 , makes it extract what was matched in the first capturing group, ([^/]+) .

Note: I'm assuming the complete value won't start or end with a slash, like /motorola products/routers or application/motorola products/ .

Answer 3

select split('application/motorola products/routers','/')[size(split('application/motorola products/routers','/'))-1]