SQL搜索中的Sum函数

GROUP BY 分组依据 SELECT 查询中的子句。在进行分组时, SDormant 还不存在。

你也不能 一个聚合表达式,因为它使人的头爆炸。

根据你的编辑,这可能会得到你想要的。它应该显示每个 Member_ID / LoanAccount_ID 一对,总共 Dormant 贷款总额 不 休眠的 贷款,以及两者的总和,只是作为一个健全的检查。

    Select  Member_Id,    
            LoanAccount_ID,
            SUM(Dormant) AS SDormant,
            SUM(CASE WHEN Dormant = 0 THEN 1 ELSE 0 END) AS NotDormant,
            COUNT(*) AS TotalCount
    From    LoanAccountBal
    Group by member_ID, LoanAccount_ID
    Order by Member_ID asc, LoanAccount_ID asc

我不知道你的问题和你的问题有什么关系。但你通常不会使用 SUM() 在 GROUP BY

Select Member_Id, LoanAccount_ID, SUM(Dormant) as SDormant
From LoanAccountBal
Group by member_ID, LoanAccount_ID
Order by Member_ID asc, LoanAccount_ID asc;

where <some column> = 1

根据你的描述,我会:

Select Member_Id, LoanAccount_ID, 
       SUM(CASE WHEN <particular field> = 1 THEN 1 ELSE 0 END) as SDormant
From LoanAccountBal
Group by member_ID, LoanAccount_ID
Order by Member_ID asc, LoanAccount_ID asc;

另一种选择是使用 COUNT() WHERE 条款:

Select Member_Id, LoanAccount_ID, COUNT(*) as SDormant
From LoanAccountBal
WHERE <particular field> = 1
Group by member_ID, LoanAccount_ID
Order by Member_ID asc, LoanAccount_ID asc;

null 如果是这样,那么你只需要 count()

Select Member_Id, LoanAccount_ID, COUNT(Dormant) as SDormant
From LoanAccountBal
Group by member_ID, LoanAccount_ID
Order by Member_ID asc, LoanAccount_ID asc;

但是,如果只修改 哪里 条款如下:

WHERE Dormant IS NOT NULL