Mysql两个聚合函数之和不起作用

嗨，我想添加两个聚合函数的结果，但是我得到了“无效使用组函数”。任何人都可以更正以下查询：

SELECT   mc.complaint_type_id,
         mc.complaint_type,
         sum(sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when ((c.is_solved = 0) and (c.res_user_id is null)) then 1 else 0 end)) as complaints_count,

    from  svk_apt_master_complaints mc
        left join svk_apt_complaints c on c.complaint_type_id = mc.complaint_type_id and c.is_active = 1
            and c.customer_id = 1 and c.association_id = 1

        group by mc.complaint_type_id

试试这个：

SELECT   mc.complaint_type_id,
         mc.complaint_type,
         sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when ((c.is_solved = 0) and (c.res_user_id is null)) then 1 else 0 end) as complaints_count
from  svk_apt_master_complaints mc
left join svk_apt_complaints c on c.complaint_type_id = mc.complaint_type_id 
where c.is_active = 1 and c.customer_id = 1 and c.association_id = 1
group by mc.complaint_type_id, mc.complaint_type

使用

+

运算符时，不需要

sum（）

。另外，

SUM

是聚合函数

SELECT   
  mc.complaint_type_id,
  mc.complaint_type,
  sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when c.is_solved = 0 and c.res_user_id is null then 1 else 0 end) as complaints_count,
FROM svk_apt_master_complaints mc
LEFT JOIN svk_apt_complaints c ON 
  c.complaint_type_id = mc.complaint_type_id 
  and c.is_active = 1
  and c.customer_id = 1 
  and c.association_id = 1
GROUP BY mc.complaint_type_id, mc.complaint_type

此外，您还可以选择未包含在聚合中的列：

mc.complaint\u type

。您需要将其包括在

分组依据中，还是将其删除。
尝试以下操作：

SELECT   mc.complaint_type_id,
         mc.complaint_type,
         sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when ((c.is_solved = 0) and (c.res_user_id is null)) then 1 else 0 end) as complaints_count
from  svk_apt_master_complaints mc
left join svk_apt_complaints c on c.complaint_type_id = mc.complaint_type_id 
where c.is_active = 1 and c.customer_id = 1 and c.association_id = 1
group by mc.complaint_type_id, mc.complaint_type

使用
+
运算符时，不需要
sum（）
。另外，
SUM
是聚合函数

SELECT   
  mc.complaint_type_id,
  mc.complaint_type,
  sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when c.is_solved = 0 and c.res_user_id is null then 1 else 0 end) as complaints_count,
FROM svk_apt_master_complaints mc
LEFT JOIN svk_apt_complaints c ON 
  c.complaint_type_id = mc.complaint_type_id 
  and c.is_active = 1
  and c.customer_id = 1 
  and c.association_id = 1
GROUP BY mc.complaint_type_id, mc.complaint_type

此外，您还可以选择未包含在聚合中的列：
mc.complaint\u type
。您需要将其包括在“分组依据”中，还是仅将其删除。
您需要在“分组依据”中指定“投诉类型”列

SELECT   mc.complaint_type_id,
         mc.complaint_type,
         sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when c.is_solved = 0 and c.res_user_id is null then 1 else 0 end) as complaints_count,
from  svk_apt_master_complaints mc
left join svk_apt_complaints c on c.complaint_type_id = mc.complaint_type_id and c.is_active = 1 and c.customer_id = 1 and c.association_id = 1
group by mc.complaint_type_id,mc.complaint_type

您需要在group by中指定mc.complaint_type列

SELECT   mc.complaint_type_id,
         mc.complaint_type,
         sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when c.is_solved = 0 and c.res_user_id is null then 1 else 0 end) as complaints_count,
from  svk_apt_master_complaints mc
left join svk_apt_complaints c on c.complaint_type_id = mc.complaint_type_id and c.is_active = 1 and c.customer_id = 1 and c.association_id = 1
group by mc.complaint_type_id,mc.complaint_type

删除嵌套的
总和
并将
mc.complaint\u type
添加到
分组依据
子句中。如果需要添加两个聚合函数的值，请使用
+
运算符，而不是聚合函数

SELECT   
  mc.complaint_type_id,
  mc.complaint_type,
  sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when c.is_solved = 0 and c.res_user_id is null then 1 else 0 end) as complaints_count,
FROM svk_apt_master_complaints mc
LEFT JOIN svk_apt_complaints c ON 
  c.complaint_type_id = mc.complaint_type_id 
  and c.is_active = 1
  and c.customer_id = 1 
  and c.association_id = 1
GROUP BY mc.complaint_type_id, mc.complaint_type

我还重新格式化了您的代码并删除了不必要的括号。
删除嵌套的
和
并将
mc.complaint\u type
添加到
分组依据
子句中。如果需要添加两个聚合函数的值，请使用
+
运算符，而不是聚合函数

SELECT   
  mc.complaint_type_id,
  mc.complaint_type,
  sum(case when c.is_solved = 1 then 1 else 0 end) + sum(case when c.is_solved = 0 and c.res_user_id is null then 1 else 0 end) as complaints_count,
FROM svk_apt_master_complaints mc
LEFT JOIN svk_apt_complaints c ON 
  c.complaint_type_id = mc.complaint_type_id 
  and c.is_active = 1
  and c.customer_id = 1 
  and c.association_id = 1
GROUP BY mc.complaint_type_id, mc.complaint_type

我还重新格式化了您的代码并删除了不必要的括号。
您不能像这样使用
SUM（SUM（…）
。你想在这里实现什么？你不能像这样使用
SUM（SUM（…）
。你在这里想要实现什么？