glorfindel-SE-Userscripts/hnq-research/HNQ-snapshots-by-site.sql

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2018-09-02 10:07:52 +00:00
SELECT temp.site, temp.percentage AS "% of HNQ", temp.questions AS "# of HNQ", t.number_of_questions AS "# of all Q on site",
ROUND(100.0 * questions / t.number_of_questions, 2) AS "% of all Q on site"
FROM (SELECT site, ROUND(100.0 * COUNT(*) / (SELECT COUNT(*) FROM snapshots), 2) AS "percentage",
COUNT(DISTINCT(question)) AS "questions"
FROM snapshots
GROUP BY site
ORDER BY COUNT(*) DESC
) AS temp
INNER JOIN total_question_count AS t on t.site = temp.site
ORDER BY 2 DESC