从 left join 获取所有用户 id

Question

我正在根据答案 table 计算分数。每个答案都有一个难度级别，分数因难度级别而异，所以我写了一个 sql 查询来计算它我成功计算了分数，但问题是当任何用户没有玩过任何一个或多个然后一个难度级别然后列即将变为空。有关更多说明，请参见下图。

我的查询是

select (IFNULL(k.l1_p*0.5,0)+IFNULL(k.l2_p*1,0)+IFNULL(k.l3_p*2,0)+IFNULL(k.l4_p*2.75,0)+IFNULL(k.l5_p*3.75,0)) as total,k.user_id from (select tab1.l1_p,tab5.user_id,tab2.l2_p,tab3.l3_p,tab4.l4_p,tab5.l5_p from (select count(id) as l1_p,user_id from wp_user_answers where answer=1 and difficulty=1 group by user_id) tab1 left join 
->         (select count(id) as l2_p,user_id from wp_user_answers where answer=1 and difficulty=2 group by user_id) tab2 on tab1.user_id=tab2.user_id left join 
->         (select count(id) as l3_p,user_id from wp_user_answers where answer=1 and difficulty=3 group by user_id) tab3  on tab3.user_id=tab2.user_id left join
->         (select count(id) as l4_p,user_id from wp_user_answers where answer=1 and difficulty=4 group by user_id) tab4  on tab3.user_id=tab4.user_id left join
->         (select count(id) as l5_p,user_id from wp_user_answers where answer=1 and difficulty=5 group by user_id) tab5  on tab4.user_id=tab5.user_id) k;

这个查询的结果是

可以看到user_id :169变成了Null。我想要所有 user_id

Answer 1

无需使用LEFT JOIN，使用聚合函数和条件可以直接得到想要的结果报表.

试试这个：

SELECT user_id, 
      (SUM(CASE WHEN difficulty = 1 THEN 1 ELSE 0 END) * 0.5 + 
       SUM(CASE WHEN difficulty = 2 THEN 1 ELSE 0 END) * 1.0 + 
       SUM(CASE WHEN difficulty = 3 THEN 1 ELSE 0 END) * 2.0 + 
       SUM(CASE WHEN difficulty = 4 THEN 1 ELSE 0 END) * 2.75 + 
       SUM(CASE WHEN difficulty = 5 THEN 1 ELSE 0 END) * 3.75 
     ) AS total
FROM wp_user_answers 
WHERE answer = 1 
GROUP BY user_id;

从 left join 获取所有用户 id

Get all user ids from left join

mysql

sql

select

aggregate-functions

left-join