Test 3 - Quarry

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

SQL

USE ckbwiki_p;
SELECT
    (SELECT COUNT(DISTINCT user_id) FROM user) AS user_count,
    (SELECT COUNT(DISTINCT up_user) FROM user_properties WHERE up_property = 'gender' AND up_value = 'male') AS male_count,
    (SELECT COUNT(DISTINCT up_user) FROM user_properties WHERE up_property = 'gender' AND up_value = 'female') AS female_count;

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