Question

在下面的查询中，它带来每个主类别中的广告计数。我还希望再增加一列，以显示所有广告的数量（无论哪个主类别，所有广告的总和）。我需要在下面的查询中带来什么变化？

谢谢

select count(*),master_category,count(a2.adid) from ad a1
inner join Category c on a1.catid=c.id
inner join ad a2 on a1.adid=a2.adid
group by c.master_category

Answer 1

如果您想添加一列总计数，则可以在按类别分类的结果和总计数之间使用交叉联接

    select  t1.master_category, t1.count_by_cat, t2.tot_count
    from  (
    select  master_category,count(a2.adid) count_by_cat 
    from ad a1
    inner join Category c on a1.catid=c.id
    inner join ad a2 on a1.adid=a2.adid
    group by c.master_category )
    ) t1
    cross join  (
        select count(*) tot_count
        inner join Category c on a1.catid=c.id
        inner join ad a2 on a1.adid=a2.adid
    ) t2

交叉联接是两组行之间的笛卡尔乘积..在这种情况下，一组用于类别结果，第二组仅用于类别结果

然后将交叉连接与总结果一起使用允许为每一行显示相同的结果..（在这种情况下，由第二个子查询计算出的总值）
通常是类别的每一行的结果与单个值

之间的联接

Answer 2

尝试这样的事情：

select c.master_category, 
       count(a1.adid) as countGroup, 
       min(a2.count1) as countAll
from ad a1
inner join Category c on a1.catid=c.id
left join (select count(ad.someColumn) as count1 from ad) as a2 on a2.adid = a1.adai
group by c.master_category

Answer 3

使用窗口功能：

select c.master_category, count(*) as category_cnt,
       sum(count(*)) over () as total_cnt
from ad a inner join
     Category c
     on a.catid = c.id
group by c.master_category;

我不明白您为什么两次加入ad。这不仅是最简单的方法，而且应该具有您想要执行的最佳性能。

Answer 4

我只是修改您的查询，希望对您有帮助

select count(
    select * from ad a1
    inner join Category c on a1.catid=c.id
    inner join ad a2 on a1.adid=a2.adid) as Total,master_category,count(a2.adid) from ad a1
    inner join Category c on a1.catid=c.id
    inner join ad a2 on a1.adid=a2.adid
    group by c.master_category

使用按计数分组查询，还可以带来广告表中所有记录的计数

4 个答案: