select s2.s_name,ifnull(round(avg(s1.s_score),1),0) as avg_score from score s1 inner join student s2 on s1.s_id = s2.s_id group by s1.s_id,s2.s_name where avg_score is null;这句sql中的错误

这句 SQL 语句中的错误是 `where` 关键字位置不正确，应该将 `where` 关键字放在 `inner join` 之后的条件中，正确的 SQL 语句如下：

select s2.s_name,ifnull(round(avg(s1.s_score),1),0) as avg_score 
from score s1 
inner join student s2 on s1.s_id = s2.s_id 
where avg_score is null
group by s1.s_id,s2.s_name;

在这个 SQL 语句中，我们使用了 `inner join` 将成绩表 `score` 中的学生 ID 与学生表 `student` 中的学生 ID 进行关联，然后使用 `group by` 对结果进行分组，最后使用 `ifnull` 函数将平均分为空的学生设置为 0。

相关问题

SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll select * FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code and P.vote_code = 'vote0000000006' SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll select * FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code and P.vote_code = 'vote0000000006' and V.app_id = 2958 GROUP BY P.option_name;错误的查出了V.vote_code 不等于 vote0000000006的数据

您好！根据您的描述，您的查询语句似乎有一个错误，导致无法正确地筛选出`V.vote_code`等于`vote0000000006`的数据。请检查您的查询语句，并确保正确使用了条件和运算符。以下是一个修正后的示例查询语句：

SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll
FROM cus_vote_option AS P
LEFT JOIN cus_vote_user AS V
    ON P.code = V.vote_option_code
WHERE P.vote_code = 'vote0000000006'
    AND V.app_id = 2958
GROUP BY P.option_name;

修正后的查询语句中，我们将筛选条件`P.vote_code = 'vote0000000006'`移动到了`WHERE`子句中，并添加了筛选条件`V.app_id = 2958`以满足您的要求。同时，我们还将`GROUP BY`子句移动到了正确的位置。请根据您的实际需求修改这个示例查询语句，以满足您的具体要求。希望对您有所帮助！

优化成mysql脚本，不使用full outer join；SELECT COALESCE(storage_in.in_date, storage_out.out_date) AS date, COALESCE(storage_in.product_name, storage_out.product_name) AS product_name, IFNULL(storage_in.quantity,0) AS IN_, IFNULL(storage_out.quantity,0) AS OUT_, SUM(IFNULL(storage_in.quantity,0) - IFNULL(storage_out.quantity,0)) OVER (PARTITION BY COALESCE(storage_in.in_date,storage_out.out_date), COALESCE(storage_in.product_name,storage_out.product_name) ORDER BY COALESCE(storage_in.in_date, storage_out.out_date)) AS END_ FROM storage_in FULL OUTER JOIN storage_out ON storage_in.product_name=storage_out.product_name AND storage_in.in_date = storage_out.out_date ORDER BY 1,2;

以下是将 FULL OUTER JOIN 转换为 MySQL 脚本的方法，使用 UNION 和 LEFT JOIN 和 RIGHT JOIN 组合：

SELECT COALESCE(storage_in.in_date, storage_out.out_date) AS date,
       COALESCE(storage_in.product_name, storage_out.product_name) AS product_name,
       IFNULL(storage_in.quantity, 0) AS IN_,
       IFNULL(storage_out.quantity, 0) AS OUT_,
       SUM(IFNULL(storage_in.quantity, 0) - IFNULL(storage_out.quantity, 0))
           OVER (PARTITION BY COALESCE(storage_in.in_date, storage_out.out_date),
                             COALESCE(storage_in.product_name, storage_out.product_name)
               ORDER BY COALESCE(storage_in.in_date, storage_out.out_date)) AS END_
FROM
  (SELECT in_date, product_name, quantity FROM storage_in
   UNION
   SELECT out_date, product_name, 0 FROM storage_out) AS storage
LEFT JOIN storage_in ON storage.in_date = storage_in.in_date
AND storage.product_name = storage_in.product_name
LEFT JOIN storage_out ON storage.out_date = storage_out.out_date
AND storage.product_name = storage_out.product_name
ORDER BY 1, 2;

这个脚本使用 UNION 将 storage_in 和 storage_out 表中的数据组合起来，LEFT JOIN 和 RIGHT JOIN 分别连接 storage_in 和 storage_out 表，从而实现 FULL OUTER JOIN 的效果。