How to pass multiple delimeters in substring_index - mysql

Question

I want to query the string between https:// or http:// and the first delimeter characters that comes after it. For example, if the field contains:
https://google.com/en/
https://www.yahoo.com?en/
I want to get:
google.com
www.yahoo.com
My initial query that will capture the / only contains two substring_index as follows:
SELECT substring_index(substring_index(mycol,'/',3),'://',-1)
FROM mytable;
Now I found that the URLs may contain multiple delimeters. I want my statament to capture multiple delimeters possibilities which are (each one is a separate character):
:/?#[]#!$&'()*+,;=
How to do this in my statement? I tried this solution but the end result the command could not be executed due to syntax error while I am sure I followed the solution. Can anyone help me correctly construct the query to capture all the delimeter characters I listed above?
I use MySQL workbecnh 6.3 on Ubuntu 18.04.
EDIT:
Some corrections made in the first example of URLs.

Answer 1

First, note that https://www.yahoo.com?en/ seems like an unlikely URL, because it has a path separator contained inside the query string. In any case, if you are using MySQL 8+, then consider using its regex functionality. The REGEXP_REPLACE function can be helpful here, using the following pattern:
https?://([A-Za-z_0-9.-]+).*
Sample query:
WITH yourTable AS (
SELECT 'https://www.yahoo.com?en/' AS url UNION ALL
SELECT 'no match'
)
SELECT
REGEXP_REPLACE(url, 'https?://([A-Za-z_0-9.-]+).*', '$1') AS url
FROM yourTable
WHERE url REGEXP 'https?://[^/]+';
Demo
The term $1 refers to the first capture group in the regex pattern. An explicit capture group is denoted by a quantity in parentheses. In this case, here is the capture group (highlighted below):
https?://([A-Za-z_0-9.-]+).*
^^^^^^^^^^^^^^^
That is, the capture group is the first portion of the URL path, including domain, subdomain, etc.

Answer 2

In MySQL 8+, this should work:
SELECT regexp_replace(regexp_substr(mycol, '://[a-zA-Z0-9_.]+[/:?]'), '[^a-zA-Z0-9_.]', '')
FROM (SELECT 'https://google.com/en' as mycol union all
SELECT 'https://www.yahoo.com?en'
) x
In older versions, this is much more challenging because there is no way to search for a string class.
One brute force method is:
select (case when substring_index(mycol, '://', -1) like '%/%'
then substring_index(substring_index(mycol, '://', -1), '/', 1)
when substring_index(mycol, '://', -1) like '%?%'
then substring_index(substring_index(mycol, '://', -1), '?', 1)
. . . -- and so on for each character
else substring_index(mycol, '://', -1)
end) as what_you_want
The [a-zA-Z0-9_.] is intended to be something like the valid character class for your domain names.