Question

我正在尝试获取所有负值的总和，所有正值的总和。这是我的查询：

SELECT
        gl.rowno,
        gl.br
        gl.fs
        gl.cudic
        gl.name
        gl.no_
        gl.balance
        SUM(glhi.amount) AS Total,
        CASE WHEN glhi.amount >0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalPositive,
        CASE WHEN glhi.amount <0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalNegative
FROM gl
        INNER JOIN glhi ON gl.rowno = glhi.rowno
WHERE 
        status = 'active'
        AND glhi.amount != 0.00
        AND glhi.effective BETWEEN '09-01-2017' AND '09-30-2017'
GROUP BY gl.rowno, gl.name, gl.no_, gl.branch, gl.fs, gl.cudic, gl.balance, glhi.effective, glhi.amount
ORDER BY gl.br, gl.name

这是我得到的当前输出的示例：

编辑：显而易见的答案是从我的GROUP BY子句中删除glhi.amount，但是当我这样做时，我得到了这个错误：

Column 'glhi.amount' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.

Answer 1

当您使用sum，avg，count等聚合函数时，似乎是正确的结果。必须对值进行分组才能进行正确的聚合

在这种情况下，您将根据此条件对结果进行分组 GROUP BY gl.rowno, gl.name, gl.no_, gl.branch, gl.fs, gl.cudic, gl.balance, glhi.effective, glhi.amount

您的字段glhi.amount在表中具有不同的值，因此此查询将为您提供多个记录

此外，请注意，在这种情况下，您应该将和放在sum函数中

CASE WHEN glhi.amount >0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalPositive,
CASE WHEN glhi.amount <0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalNegative

更改为

sum(case when glhi.amount > 0 then coalesce(glhi.amount, 0)) end as TotalPositive
sum(case when glhi.amount < 0 then coalesce(glhi.amount, 0)) end as TotalNegative

Answer 2

根据输出数据，您需要删除glhi.amount中的GROUP BY

SELECT
        gl.rowno,
        gl.br
        gl.fs
        gl.cudic
        gl.name
        gl.no_
        gl.balance
        SUM(glhi.amount) AS Total,
        CASE WHEN glhi.amount >0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalPositive,
        CASE WHEN glhi.amount <0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalNegative
    FROM gl
        INNER JOIN glhi ON gl.rowno = glhi.rowno
    WHERE 
        status = 'active'
        AND glhi.amount != 0.00
        AND glhi.effective BETWEEN '09-01-2017' AND '09-30-2017'
        GROUP BY gl.rowno, gl.name, gl.no_, gl.branch, gl.fs, gl.cudic, gl.balance, glhi.effective
        ORDER BY gl.br, gl.name

Answer 3

我认为最好的逻辑是：

sum(case when glhi.amount > 0 then glhi.amount else 0 end) as TotalPositive
sum(case when glhi.amount < 0 then glhi.amount else 0 end) as TotalNegative

这称为“条件聚集”。 NULL比较不是必需的； else 0可以解决这个问题。

Answer 4

如果每个其他项目有多个glhi.amount，它将返回多行。尝试像这样删除glhi.amount：

SELECT
        gl.rowno,
        gl.br
        gl.fs
        gl.cudic
        gl.name
        gl.no_
        gl.balance
        SUM(glhi.amount) AS Total,
        CASE WHEN glhi.amount >0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalPositive,
        CASE WHEN glhi.amount <0 THEN ISNULL(SUM(glhi.amount),0) END AS TotalNegative
    FROM gl
        INNER JOIN glhi ON gl.rowno = glhi.rowno
    WHERE 
        status = 'active'
        AND glhi.amount != 0.00
        AND glhi.effective BETWEEN '09-01-2017' AND '09-30-2017'
        GROUP BY gl.rowno, gl.name, gl.no_, gl.branch, gl.fs, gl.cudic, gl.balance, glhi.effective
        ORDER BY gl.br, gl.name

Answer 5

处理2个案例并将其放在SUM（）子句中，然后可以删除GROUP BY金额并生效。谢谢大家的帮助

select DISTINCT
    gl.rowno,
    gl.br
    gl.fs 
    gl.cudic,
    gl.name
    gl.no_
    gl.balance 
    SUM(glhi.amount) AS Total,
    SUM(CASE WHEN glhi.amount >0 THEN glhi.amount END)AS TotalPositive,
    SUM(CASE WHEN glhi.amount <0 THEN glhi.amount END)AS TotalNegative
FROM gl
    INNER JOIN glhi ON gl.rowno = glhi.rowno
WHERE 
    status = 'active'
    AND glhi.amount != 0.00
    AND glhi.effective BETWEEN '09-01-2017' AND '09-30-2017'
    GROUP BY gl.rowno, gl.name, gl.no_, gl.br, gl.fs, gl.cudic, gl.balance
ORDER BY branch, name

sum（）返回多行

5 个答案: