Case 语句给出 group by 子句错误

Question

以下查询：

SELECT 
    CASE WHEN is_dog = 1 THEN COUNT(*) END AS dogs,
    CASE WHEN is_cats = 1 THEN count(*) END as cats
FROM 
    animals
WHERE
    birth_date >= '2018-01-01';

我收到这个错误：

Invalid operation: column "is_dog" must appear in the GROUP BY clause or be used in an aggregate function

我需要对此进行哪些更改？我刚开始使用 redshift。

Answer 1

应该写成

SELECT count(*) FILTER (WHERE is_dog = 1) AS dogs,
       count(*) FILTER (WHERE is_cat = 1) AS cats
FROM animals
WHERE birth_date >= '2018-01-01';

Answer 2

在 Amazon Redshift 上，您可能需要：

SELECT SUM(case when is_dog = 1 then 1 else 0 END) AS dogs,
       SUM(case when is_cats = 1 then then 1 else 0 end)  as cats
FROM animals
WHERE birth_date >= '2018-01-01';

或者，更短的形式是：

SELECT SUM( (is_dog = 1)::int ) AS dogs,
       SUM( (is_cats = 1)::int )  as cats
FROM animals
WHERE birth_date >= '2018-01-01';

或者如果列只取0/1/NULL，那么：

SELECT SUM( is_dog ) AS dogs,
       SUM( is_cats )  as cats
FROM animals
WHERE birth_date >= '2018-01-01';

Answer 3

COUNT(*) 表达式在不同的列或不同的 CASE 表达式中不会不同。是行数，在外加上其他条件不会改变

此外，当您将所有行分组为一行时，您不能引用 is_cat 或 is_dog，除非该引用在聚合函数内 (例如 SUM() 或 COUNT()).

将这两个事实放在一起，您可以做...

SUM(CASE WHEN is_dog = 1 THEN 1 ELSE 0 END)

这给出了类似...的查询

SELECT SUM(CAST WHEN is_dog = 1 THEN 1 ELSE 0 END) AS dogs, SUM(CAST WHEN is_cat = 1 THEN 1 ELSE 0 END) AS cats FROM animals WHERE birth_date >= '2018-01-01';

或者，假设 is_cat 和 is_dog 字段只有 0 或 1，您实际上甚至不需要 CASE 表达式所有...

SELECT SUM(is_dog) AS dogs, SUM(is_cat) AS cats FROM animals WHERE birth_date >= '2018-01-01' ;

Case 语句给出 group by 子句错误

Case statement gives group by clause error

sql

conditional

amazon-redshift