汤姆Kyte
suggests使用
EXTRACT得到的区别：
extract( day from (x-y) )*24*60*60+ extract( hour from (x-y) )*60*60+ ...
这似乎比这更难读,比较慢：
( CAST( x AS DATE ) - CAST( y AS DATE ) ) * 86400
那么,在几秒钟内得到两个时间戳之间的区别呢？谢谢！
“最佳实践”
原文链接：https://www.f2er.com/oracle/205545.html无论你做什么,把它包装在一个功能中,例如seconds_between(from_date,to_date) – 无论它如何操作(选择最有效的方法) – 那么你的代码将是非常明显的.
我在笔记本电脑(WinXP)上测试了11gR1上的两种方法,下面是测试用例.看来CAST选项是最快的. (t1是基线,t2使用提取方法,t3使用cast方法)
t1 (nothing) 3 t2 (extract) 338 t3 (cast) 101 t1 (nothing) 3 t2 (extract) 336 t3 (cast) 100
测试脚本
declare
x TIMESTAMP := SYSTIMESTAMP;
y TIMESTAMP := TRUNC(SYSDATE);
n PLS_INTEGER;
lc CONSTANT PLS_INTEGER := 1000000;
t1 PLS_INTEGER;
t2 PLS_INTEGER;
t3 PLS_INTEGER;
begin
t1 := DBMS_UTILITY.get_time;
for i in 1..lc loop
n := i;
end loop;
t1 := DBMS_UTILITY.get_time - t1;
t2 := DBMS_UTILITY.get_time;
for i in 1..lc loop
n := extract(day from (x-y))*24*60*60
+ extract(hour from (x-y))*60*60
+ extract(minute from (x-y))*60
+ extract(second from (x-y));
end loop;
t2 := DBMS_UTILITY.get_time - t2;
t3 := DBMS_UTILITY.get_time;
for i in 1..lc loop
n := ( CAST( x AS DATE ) - CAST( y AS DATE ) ) * 86400;
end loop;
t3 := DBMS_UTILITY.get_time - t3;
dbms_output.put_line('t1 (nothing) ' || t1);
dbms_output.put_line('t2 (extract) ' || t2);
dbms_output.put_line('t3 (cast) ' || t3);
end;
