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 p.up_user END) AS male_count,
    COUNT(DISTINCT CASE WHEN p.up_value = 'female' THEN p.up_user END) AS female_count,
    COUNT(DISTINCT u.user_id) - COUNT(DISTINCT CASE WHEN p.up_value IN ('male', 'female') THEN u.user_id END) AS remaining_count
FROM
    user AS u
LEFT JOIN
    user_properties AS p ON u.user_id = p.up_user AND p.up_property = 'gender';

By running queries you agree to the Cloud Services Terms of Use and you irrevocably agree to release your SQL under CC0 License.

Submit Query Stop Query

All SQL code is licensed under CC0 License.

SQL

Checking query status...