Question

我有一张数据如下的表：

Meal ID   |   Child    |  Fruit

1              Child A     Apple
2              Child B     Orange
3              Child A     Banana
4              Child A     Orange
5              Child C     Banana
6              Child A     Banana
7              Child C     Banana
8              Child A     Banana

，我需要根据所吃的水果（例如如果搜索查询是“香蕉”，结果将显示如下：

Banana Eaters (Highest to Lowest)

Child   |   Number of Bananas eaten

Child A     3
Child C     1

我将从：

开始

 $sql = "SELECT * FROM meals WHERE Fruit = 'Banana'";

但是我该如何将吃香蕉的独特个体作为自己的条目，然后按食用最多的香蕉排序结果？

Answer 1

按孩子分组并使用条件聚合：

select child, sum(fruit = 'Banana') `Number of Bananas eaten`
from meals
group by child
order by `Number of Bananas eaten` desc

如果要排除没有香蕉的孩子：

select child, sum(fruit = 'Banana') `Number of Bananas eaten`
from meals
group by child
having `Number of Bananas eaten` > 0
order by `Number of Bananas eaten` desc

请参见demo。

Answer 2

用例何时

select child,sum(case when Fruit='Banana' then 1 else 0 end) as bananaeaten
from meals
group by child
order by bananaeaten desc

Answer 3

select Child, count(*) as 'Number of bananas eaten' from meals
where Fruit = 'Banana'
group by Child
order by count(*) desc

Answer 4

如果您不想看到“孩子B”，请在汇总之前进行过滤：

select Child, count(*) as num_bananas
from meals
where Fruit = 'Banana'
group by Child
order by count(*) desc;

SQL按结果计数的顺序

4 个答案: