一：
select count(*) as 总数 from A .....
union all
select count(*) as 有效数 from A where 条件='有效'
union all
select count(*) as 总数 from B .....
union all
select count(*) as 有效数 from B where 条件='有效'
union all
select count(*) as 总数 from C .....
union all
select count(*) as 有效数 from C where 条件='有效'
union all
select count(*) as 总数 from D .....
union all
select count(*) as 有效数 from D where 条件='有效'
union all
select count(*) as 总数 from E .....
union all
select count(*) as 有效数 from E where 条件='有效'

二：
select "A", count(*) as 总数, sum(if(条件='有效', 1, 0) as 有效数 from A
union 
select "B", count(*) as 总数, sum(if(条件='有效', 1, 0) as 有效数 from B
union 
select "C", count(*) as 总数, sum(if(条件='有效', 1, 0) as 有效数 from C
union 
select "D", count(*) as 总数, sum(if(条件='有效', 1, 0) as 有效数 from D
union 
select "E", count(*) as 总数, sum(if(条件='有效', 1, 0) as 有效数 from E

------解决方案--------------------
你就直接UNION吧。

------解决方案--------------------
直接union all性能会好一些的。

------解决方案--------------------
union all就可以的。