MySQL和数量

现在我有一个问题：

SELECT 
    Categoria.Descricao AS Categoria,
    Marca.descricao AS Marca,
    Modelo.descricao AS Modelo,Material.n_serie As 'Nº Série', 
    SUM(Entrada.Qtd-COALESCE(Saida.Qtd,0)) AS QTD ,username AS 'User' 
FROM Marca,Modelo,Categoria,Material,Users,Entrada 
    LEFT JOIN Saida ON Entrada.n_serie=Saida.n_serie 
WHERE Marca.id_marca=Modelo.id_marca 
    AND Modelo.categoria= Categoria.id_categoria 
    AND Modelo.id_modelo=Material.id_modelo 
    AND Material.n_serie=Entrada.n_serie 
    AND Entrada.user=Users.id 
GROUP BY Categoria.descricao,Marca.descricao,Modelo.descricao,Material.n_serie;

输出

 Categoria, Marca, Modelo,      Nº Série, QTD,  User
'Cabo',     'UTP', '3 metros', '1234',    '2', 'admin'

正如你所见，它告诉我我还有两条UTP电缆，而我只有一条

表格Entrada（In）：

 n_serie, data_entrada, user, qtd, obs
'1234', '2014-05-12 14:37:34', '6', '50', ''
'1234', '2014-05-12 14:37:43', '6', '2', ''

 n_serie, data_saida, user, qtd, obs
'1234', '2014-05-12 14:50:16', '6', '2', ''
'1234', '2014-05-12 14:50:22', '6', '49', ''

50+2=52

赛达表（外）：

 n_serie, data_entrada, user, qtd, obs
'1234', '2014-05-12 14:37:34', '6', '50', ''
'1234', '2014-05-12 14:37:43', '6', '2', ''

 n_serie, data_saida, user, qtd, obs
'1234', '2014-05-12 14:50:16', '6', '2', ''
'1234', '2014-05-12 14:50:22', '6', '49', ''

49+2=51

有人能告诉我这个问题在哪里找到另一个问题吗

---

提前感谢。

问题是，您将2行与2行合并，结果得到4行，然后将4行相加

（Entrada中的每一行都与Saida中的两行匹配。）

您需要做的是先聚合每个表，然后将它们连接在一起

(
  SELECT n_serie, SUM(qtd) AS qtd
    FROM Entrada
GROUP BY n_serie
)
  AS entrada
LEFT JOIN
(
  SELECT n_serie, SUM(qtd) AS qtd
    FROM Saida
GROUP BY n_serie
)
  AS Saida
    ON Saida.n_serie = Entrada.n_serie

这会给你

Entrada:         Saida:

n_serie, qtd     n_serie, qtd        a - b
'1234' , 52      '1234' , 51           1

---

您混合了隐式和显式

join

语法。隐式

join

语法（条件在

where

子句中）是危险的——很容易出错。我建议您使用显式联接重写查询，答案可能很明显。