如何劃分兩個不同的查詢數（）在DB2

Question

所以我

select count(*) from ("query1") 

select count(*) from ("query2") 

我想分兩個並獲得浮點結果。

我被告知要使用這樣的

SELECT (COUNT(smtg) * 1.0)/COUNT(smtg)

但林不知道

Answer 1

你可以這樣做：

select q1.cnt * 1.0/q2.cnt 
from (select count(*) as cnt from ("query1")) q1 cross join 
    (select count(*) as cnt from ("query2")) q2; 

或者，如果你喜歡：

select ((select count(*) from ("query1")) * 1.0/
     (select count(*) from ("query2")) 
     ) 
from sysibm.sysdummy1; 

Answer 2

其他解決方案（小心不要除以0）

with 
query1 as (select count(*) as nb1 from ("query1")), 
query2 as (select count(*) as nb2 from ("query2")) 

select case when nb2=0 then null else nb1* 1.0/nb2 end as Result 
from query1, query2