Test 3 - Quarry

This query is marked as a draft This query has been published by Aram.

SQL

USE ckbwiki_p;
SELECT
    COUNT(DISTINCT u.user_id) AS user_count,
    COUNT(DISTINCT CASE WHEN p.up_value = 'male' THEN u.user_id END) AS male_count,
    COUNT(DISTINCT CASE WHEN p.up_value = 'female' THEN u.user_id END) AS female_count,
    COUNT(DISTINCT CASE WHEN p.up_value NOT IN ('male', 'female') OR p.up_value IS NULL THEN u.user_id END) AS remaining_count
FROM
    user u
LEFT JOIN
    user_properties p ON u.user_id = p.up_user AND p.up_property = 'gender';

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