如果你想所有警报的数量,从数M的的C和编号,然后
SELECT
user_name,
COUNT(type_alertes) AS nb,
id_user,
SUM(CASE WHEN TYPE ='C' THEN 1 ELSE 0 END) AS critique,
SUM(CASE WHEN TYPE ='M' THEN 1 ELSE 0 END) AS majeur
FROM stat_alert
JOIN user_qdf ON user_qdf.`id` = id_user
GROUP BY id_user, user_name
应该这样做。 我假设user_name来自user_qdf表,其他来自stat_alert并且id_user不为null。
编辑:对于alertes
的总和
SELECT
user_name,
COUNT(type_alertes) AS nb,
id_user,
SUM(CASE WHEN TYPE ='C' THEN 1 ELSE 0 END) AS critique,
SUM(CASE WHEN TYPE ='M' THEN 1 ELSE 0 END) AS majeur
SUM(CASE WHEN TYPE ='M' OR TYPE = 'C' THEN 1 ELSE 0 END) AS both_c_and_m
FROM stat_alert
JOIN user_qdf ON user_qdf.`id` = id_user
GROUP BY id_user, user_name
EDIT2:
SELECT
user_name,
COUNT(type_alertes) AS nb,
SUM(CASE WHEN TYPE ='C' THEN type_alertes ELSE 0 END) AS critique_alerts,
SUM(CASE WHEN TYPE ='M' THEN type_alertes ELSE 0 END) AS majeur_alertes
id_user,
SUM(CASE WHEN TYPE ='C' THEN 1 ELSE 0 END) AS critique,
SUM(CASE WHEN TYPE ='M' THEN 1 ELSE 0 END) AS majeur
FROM stat_alert
JOIN user_qdf ON user_qdf.`id` = id_user
WHERE TYPE in ('C','M')
GROUP BY id_user, user_name
是你想要的吗?
由 – Kashif
相同的结果 – user1794019
你为什么要加入同一个表多次添加型,USER_NAME组?我怀疑你的加入。表user_qdf和id_user之间的连接字段在哪里 – polin