mysql – 在单个查询中进行分组,排序和计数
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我正在尝试GROUP,SORT和COUNT在我的一个名为’商品’的表中的一个查询中.
这是我的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
我想要 :
>按名称分组
>按家庭,次家庭和最后名称排序
> COUNT每个系列的行数,sub_family然后命名(在同一个sub_family)
到目前为止,我设法在同一个sub_family中执行除COUNT之外的所有操作.
的确,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
这导致以下结果:
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 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计算柠檬2次,但我希望它只计算一次因为一个柠檬在Fruits sub_family而另一个柠檬在Tree sub_family.
[更新]:这是我想要的结果:
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
解决方法:
只是我猜你要求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
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