我有几张table：
m_foldercommittee：
| id_folder| id_committe|意见|
| --|--|--|
| 1 | 1 | 1 |
| 2 | 2 | 1 |
| 3 | 3 | 1 |
m_folder：
| id_folder| id_committe|
| --|--|
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
m_schedule：
| id_schedule|类型表|id_folder|
| --|--|--|
| 1 | 1 | 1 |
| 2 | 2 | 2 |
| 3 | 3 | 3 |
m_committee：
| 身份证委员会|日期委员会|
| --|--|
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
现在，查询看起来像这样：

select c.date_committee,
    sum(1) as total,
    sum(if(fc.opinion= 1 , 1, 0)) as nb_folder,
    sum(if(fc.opinion= 2 , 1, 0)) as nb_folder_favorable,
    sum(if(fc.opinion= 3 , 1, 0)) as nb_folder_principe,
from m_foldercommittee fc
    INNER JOIN m_committee c ON c.id_committee = fc.id_committee
    INNER JOIN m_folder mf ON mf.id_folder = fc.id_folder
where EXISTS(SELECT NULL 
             FROM m_schedule sch 
             WHERE sch.type_schedule = 1 
             AND sch.id_folder = fc.id_folder  
             AND ((sch.beginning_date >= '2022-08-25' AND sch.ending_date <= '2023-09-25')
               OR (sch.beginning_date <  '2022-08-25' AND sch.ending_date >= '2022-08-25')
               OR (sch.beginning_date <= '2023-09-25' AND sch.ending_date >= '2023-09-25')
             ))
group by fc.id_committee

字符串
并返回类似这样的内容：
| 日期委员会|总|nb_folder| nb_folder_favorable| nb_folder_principe|
| --|--|--|--|--|
| 2023-08-25 2023-08-25 2023-08-25| 13 | 12 | 1 | 0 |
| 2023-09-15 2023-09-15| 24 | 20 | 4 | 0 |
| 2023-09-25 2023-09-25| 91 | 88 | 3 | 0 |
我想要的是插入一个新列，它是type_schedule = 1的sum，但我不想成为sum（1）的一部分。
以下是我到目前为止写的：

select c.date_committee,
    sum(1) as total,
    sum(if(sch.type_schedule= 1 , 1, 0)) as nb_beverage,
    sum(if(fc.opinion= 1 , 1, 0)) as nb_folder,
    sum(if(fc.opinion= 2 , 1, 0)) as nb_folder_favorable,
    sum(if(fc.opinion= 3 , 1, 0)) as nb_folder_principe,
from m_foldercommittee fc
    INNER JOIN m_committee c ON c.id_committee = fc.id_committee
    INNER JOIN m_folder mf ON mf.id_folder = fc.id_folder
    INNER JOIN m_schedule sch ON sch.id_folder = fc.id_folder
where EXISTS(SELECT NULL 
             FROM m_schedule sch 
             WHERE sch.type_schedule = 1 and sch.id_folder = fc.id_folder  
             AND ((sch.beginning_date >= '2022-08-25' AND sch.ending_date <= '2023-09-25')
               OR (sch.beginning_date <  '2022-08-25' AND sch.ending_date >= '2022-08-25')
               OR (sch.beginning_date <= '2023-09-25' AND sch.ending_date >= '2023-09-25')
             ))
group by fc.id_committee

型
它的工作，但问题是，第二个版本的查询返回值是完全不同的第一个版本的查询.如何排除的结果sum(if(sch.type_schedule= 1 , 1, 0)) as nb_beverage从总和（1），所以新版本的查询返回完全相同的结果作为第一个，但一个新的列 nb_beverage？