php mysql：计算其他表中不存在的元素数

我现在有这种桌子

table a
==============
pid |   block
--------------
1   |   1
2   |   1
3   |   2
4   |   3
5   |   2
6   |   3
--------------

table b
==============
cid |   pid
--------------
1   |   1
2   |   3
3   |   5
--------------

output (count each block who no exist on table b)
===================
block   |   total
-------------------
1       |   1 <--- because on table b exist 1 from 2
2       |   0 <--- because on table b exist 2 from 2
3       |   2 <--- because on table b not exist
-------------------

表a
==============
pid |块
--------------
1   |   1
2   |   1
3   |   2
4   |   3
5   |   2
6   |   3
--------------
表b
==============
cid | pid
--------------
1   |   1
2   |   3
3   |   5
--------------
输出（统计表b中不存在的每个块）
===================
区块|总计
-------------------
1 | 1您可以使用以下SQL：

select outera.block , count(outera.block)- (select count(b.pid) from a left join b on a.pid=b.pid  where a.block=outera.block
 group by a.block) as acount  
from a as outera group by outera.block;

请让我知道它是否有效。
谢谢。
为什么块id 1有计数1，你能详细解释一下吗？你可以在表a上看到块1存在2次，在pid 1和pid 2上，同时在表b上，块1只添加了1次，即pid 1sorry@holgrich，我取消了接受的答案，因为我在表a上添加另一个数据时得到的结果不正确，例如，pid 8的值为3，3的结果仍然为2，假设为3，这里是小提琴
select outera.block , count(outera.block)- (select count(b.pid) from a left join b on a.pid=b.pid  where a.block=outera.block
 group by a.block) as acount  
from a as outera group by outera.block;