如何检查两次之间的差异是否大于8.30？

Question

SELECT to_char(TO_CHAR(T1_TO_TIME,'HH:MI')-TO_CHAR(T1_FR_TIME,'HH:MI')) 
FROM Table1 

Answer 1

 
8.5 hours/24 hours = 0.3541666 of a day. 

In Oracle, one way to do it. 

select 
    case 
    when (to_date('6-4-2015 17:10','dd-mm-yyyy hh24:mi') 
     -to_date('6-4-2015 8:45','dd-mm-yyyy hh24:mi')) 
     > 0.3541666 then 
     'More than 8.30 hours' 
    else 
     'Less than 8.30 hours' 
    end 
from dual