MySQL-按天计算的组比较计数

我有以下疑问：

select
    DATE_FORMAT(created_at,'%d %M') as day,
    count(*) as Total,
    count(content_sentiment = 0) as Neutral,
    count(content_sentiment < 0) as Negative,
    count(content_sentiment > 0) as Positive
from Master
where content_sentiment IS NOT NULL
GROUP BY DAY(created_at)
;

总值应该是正值、中性值和负值的总和，这不是猜测，因为每个表的查询都会得到整个表的总值，而不是按天分组

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如何为每个正值、中性值和负值包含group by？

您需要条件聚合：

...
count(*) as Total,
SUM(CASE WHEN content_sentiment = 0 THEN 1 ELSE 0 END) as Neutral,
SUM(CASE WHEN content_sentiment < 0 THEN 1 ELSE 0 END) as Negative,
SUM(CASE WHEN content_sentiment > 0 THEN 1 ELSE 0 END) as Positive
...

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返回SELECT语句检索到的行中expr的非空值数。谢谢，非常有用。

...
count(*) as Total,
SUM(CASE WHEN content_sentiment = 0 THEN 1 ELSE 0 END) as Neutral,
SUM(CASE WHEN content_sentiment < 0 THEN 1 ELSE 0 END) as Negative,
SUM(CASE WHEN content_sentiment > 0 THEN 1 ELSE 0 END) as Positive
...

    select
    DATE_FORMAT(created_at,'%d %M') as day,
    count(*) as Total,
    sum(if(content_sentiment = 0,1,0)) as Neutral,
    sum(if(content_sentiment < 0,1,0)) as Negative,
    sum(if(content_sentiment > 0,1,0)) as Positive
from Master
where content_sentiment IS NOT NULL
GROUP BY DAY(created_at)
;