Мне удалось выяснить это, я надеюсь, что это пригодится кому-то в будущем:
SELECT
Browser,
Version,
COUNT(Browser) AS Count
FROM
(
SELECT
CASE
WHEN userAgent LIKE '%Firefox%' THEN 'Firefox'
WHEN userAgent LIKE '%Chrome%' THEN 'Chrome'
WHEN userAgent LIKE '%MSIE %' THEN 'IE'
ELSE 'Unknown'
END AS Browser,
CASE
WHEN userAgent LIKE '%Firefox%' THEN SUBSTRING(userAgent, LOCATE('Firefox', userAgent) + 9, POSITION('.' IN SUBSTRING(userAgent, LOCATE('Firefox', userAgent) + 9)) + 1)
WHEN userAgent LIKE '%Chrome%' THEN SUBSTRING(userAgent, LOCATE('Chrome', userAgent) + 8, POSITION('.' IN SUBSTRING(userAgent, LOCATE('Chrome', userAgent) + 8)) + 1)
WHEN userAgent LIKE '%MSIE %' THEN SUBSTRING(userAgent, LOCATE('MSIE ', userAgent) + 5, POSITION('.' IN SUBSTRING(userAgent, LOCATE('MSIE ', userAgent) + 5)) + 1)
ELSE 'Unknown'
END AS Version
FROM user_log
)
AS Browsers
GROUP BY Browser, Version