SQL Regex - Select everything after '/' and split into array

Question

I have to write a HSQLDB query that splits this string on '/'

/2225/golf drive/#305/Huntsville/AL/1243

This is where I am at

select REGEXP_SUBSTRING_ARRAY(Terms, ''/[a-zA-Z0-9]*'') as ARR from Address

This is giving me

/2225, /golf, /, /Huntsville, /AL, /1243 - (Missing "#305" and "drive" in second split)

How can I modify the regex such that it includes everything after "/" and give me this result

/2225, /golf drive, /#305, /Huntsville, /AL, /1243

Denniselite · Accepted Answer · 2017-09-08 22:34:47Z

4

In this case why can't you use /[a-zA-Z0-9, #]* regexp? It seems good for your goal.

I've checked, it works here for me: https://regex101.com/r/8bJQEk/1

PS This regexp /\/([^\/]*)/g can helps to split everything. Be careful with slashes). Example

answered Sep 8, 2017 at 22:20

Denniselite

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3 Comments

But this only works with #. I can get any special character in the string

@keanu If you want to split anything you can use it /([^/]*) regexp. It works here

Simple and elegant. Thanks Denniselite