Question

我想选择已注册参加某项活动的人数，该费用由signup_limit提供，存储在activity_info中，以及已注册该活动的人数，存储在signups中。如果没有人注册，我希望它以注册数量计算为零，但是正如我所知，它会完全丢弃该行。我试图使用左外连接，但我得到相同的结果。有什么建议吗？

SELECT info.code, info.signup_limit - COUNT(*) AS swag
     FROM activity_info AS info 
         LEFT OUTER JOIN signups ON info.code = signups.activity_code, variables
WHERE 
    info.code IN (
      SELECT schedule.code FROM schedule,variables WHERE schedule.date = variables.week
    ) AND variables.week = signups.date
GROUP BY signups.activity_code

更新：感谢评论解决了这个问题。谢谢大家！

Answer 1

SQL变量中间有一个单词。移除以保持良好状态：

SELECT info.code,
       info.signup_limit - count(*) AS swag
FROM activity_info AS info
LEFT OUTER JOIN signups ON info.code = signups.activity_code
WHERE info.code IN
    (SELECT schedule.code
     FROM schedule
     INNER JOIN variables ON schedule.date = variables.week)
  AND variables.week = signups.date
GROUP BY signups.activity_code

Answer 2

我想这就是你想要的：

SELECT 
    AI.code, 
    (
        AI.signup_limit - 
        (
            SELECT 
                COUNT(*)
            FROM 
                signups AS S
            WHERE 
                S.activity_code = AI.code
        )
    ) AS swag
FROM 
    activity_info AS AI
WHERE
    info.code IN 
    (
        SELECT
            schedule.code 
        FROM 
            schedule INNER JOIN variables ON schedule.date = variables.week
    ) AND 
    variables.week = signups.date

SQL查询左连接问题

2 个答案: