在单个查询中进行分组、排序和计数
在单个查询中进行分组、排序和计数
提问于 2015-04-16 16:25:51
我正在尝试在一个查询中GROUP、SORT和COUNT --我的一个表名为“初级商品”。
下面是我的MySql表的简化：
family sub_family name detailed_name
Agro Grains Wheat Wheat per 1 mt
Agro Grains Corn Corn per 1 mt
Agro Grains Sugar Sugar per 1 mt
Agro Fruits Apple Apple red
Agro Fruits Apple Apple green
Agro Fruits Apple Apple yellow
Agro Fruits Lemon Lemon classic
Wood Tree Lemon Lemon in logs
Wood Tree Oak Oak in logs
Wood Tree Epicea Epicea in logs
Wood Packaging Kraftliner Krafliner 3mm我想：
GROUPbynameSORT由family，sub_family和最后nameCOUNT每个family、sub_family和name的行数(在同一个sub_family中)
到目前为止，除了COUNT之外，我还在同一个sub_family中完成了所有工作。
实际上，以下查询：
SELECT
TableC.family,
TableC.NbrFamily,
TableB.sub_family,
TableB.NbrSubFamily,
TableA.name,
TableA.NbrName
FROM
(
SELECT
family,
sub_family,
name,
COUNT(DISTINCT commodities.id) AS NbrName
FROM commodities GROUP BY name
) TableA
INNER JOIN
(
SELECT
sub_family,
COUNT(DISTINCT commodities.id) AS NbrSubFamily
FROM commodities GROUP BY sub_family
) TableB
ON (TableA.sub_family = TableB.sub_family)
INNER JOIN
(
SELECT
family,
COUNT(DISTINCT commodities.id) AS NbrFamily
FROM commodities GROUP BY family
) TableC
ON (TableA.family = TableC.family)
GROUP BY TableA.name
ORDER BY TableA.family,TableA.sub_family,TableA.name其结果如下：
family NbrFamily sub_family NbrSubFamily name NbrName
Agro 7 Grains 3 Wheat 1
Agro 7 Grains 3 Corn 1
Agro 7 Grains 3 Sugar 1
Agro 7 Fruits 4 Apple 3
Agro 7 Fruits 4 Lemon 2
Wood 4 Tree 3 Lemon 2
Wood 4 Tree 3 Oak 1
Wood 4 Tree 3 Epicea 1
Wood 4 Packaging 1 Kraftliner 1您可以看到，NbrName计数柠檬2倍，但我希望它只数1次，因为一个柠檬在水果sub_family和另一个在树sub_family。
更新:以下是我想要的结果：
family NbrFamily sub_family NbrSubFamily name NbrName
Agro 7 Grains 3 Wheat 1
Agro 7 Grains 3 Corn 1
Agro 7 Grains 3 Sugar 1
Agro 7 Fruits 4 Apple 3
Agro 7 Fruits 4 Lemon 1
Wood 4 Tree 3 Lemon 1
Wood 4 Tree 3 Oak 1
Wood 4 Tree 3 Epicea 1
Wood 4 Packaging 1 Kraftliner 1回答 1
Stack Overflow用户
回答已采纳
发布于 2015-04-16 17:20:09
我猜你要的是什么http://sqlfiddle.com/#!9/e9206/16
因为它带来了预期的结果：
SELECT A.family, C.NbrFamily,A.sub_family,B.NbrSubFamily,A.name,COUNT(A.Name)
FROM commodities as A
LEFT JOIN (
SELECT family,sub_family,COUNT(Name) AS NbrSubFamily
FROM commodities
GROUP BY family,sub_family
) B
ON A.sub_family = B.sub_family
AND A.family = B.family
LEFT JOIN (
SELECT family,COUNT(Name) AS NbrFamily
FROM commodities
GROUP BY family
) C
ON A.family = C.family
GROUP BY A.family,A.sub_family,A.name
ORDER BY A.id页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/29680707
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