如何计算 Oracle 中两个时间戳之间的时间差（以毫秒为单位）？
【问题讨论】：
DATE 只有秒级的精度。 Oracle TIMESTAMP 具有亚秒级精度——通常是毫秒或微秒，具体取决于平台。
我知道很多人认为这个解决方案简单明了：
create table diff_timestamp (
f1 timestamp
, f2 timestamp);
insert into diff_timestamp values(systimestamp-1, systimestamp+2);
commit;
select cast(f2 as date) - cast(f1 as date) from diff_timestamp;
宾果游戏！
【讨论】：
I) 如果您需要计算两个时间戳列之间经过的时间（以秒为单位），请尝试以下操作：
SELECT
extract ( day from (end_timestamp - start_timestamp) )*86400
+ extract ( hour from (end_timestamp - start_timestamp) )*3600
+ extract ( minute from (end_timestamp - start_timestamp) )*60
+ extract ( second from (end_timestamp - start_timestamp) )
FROM table_name
II) 如果您只想以字符格式显示时差，请尝试以下操作：
SELECT to_char (end_timestamp - start_timestamp) FROM table_name
【讨论】：
我已经发布了here 一些将时间间隔转换为纳秒和纳秒转换为时间间隔的方法。这些方法具有纳秒级精度。
您只需将其调整为毫秒而不是纳秒。
一种将时间间隔转换为纳秒的更短方法。
SELECT (EXTRACT(DAY FROM (
INTERVAL '+18500 09:33:47.263027' DAY(5) TO SECOND --Replace line with desired interval --Maximum value: INTERVAL '+694444 10:39:59.999999999' DAY(6) TO SECOND(9) or up to 3871 year
) * 24 * 60) * 60 + EXTRACT(SECOND FROM (
INTERVAL '+18500 09:33:47.263027' DAY(5) TO SECOND --Replace line with desired interval
))) * 100 AS MILLIS FROM DUAL;
MILLIS
1598434427263.027
【讨论】：
Select date1 - (date2 - 1) * 24 * 60 *60 * 1000 from Table;
【讨论】：
我知道这已被详尽地回答，但我想与大家分享我的功能。它使您可以选择是否希望您的答案以天、小时、分钟、秒或毫秒为单位。您可以对其进行修改以满足您的需要。
CREATE OR REPLACE FUNCTION Return_Elapsed_Time (start_ IN TIMESTAMP, end_ IN TIMESTAMP DEFAULT SYSTIMESTAMP, syntax_ IN NUMBER DEFAULT NULL) RETURN VARCHAR2 IS
FUNCTION Core (start_ IN TIMESTAMP, end_ IN TIMESTAMP DEFAULT SYSTIMESTAMP, syntax_ IN NUMBER DEFAULT NULL) RETURN VARCHAR2 IS
day_ VARCHAR2(7); /* This means this FUNCTION only supports up to 99 days */
hour_ VARCHAR2(9); /* This means this FUNCTION only supports up to 999 hours, which is over 41 days */
minute_ VARCHAR2(12); /* This means this FUNCTION only supports up to 9999 minutes, which is over 17 days */
second_ VARCHAR2(18); /* This means this FUNCTION only supports up to 999999 seconds, which is over 11 days */
msecond_ VARCHAR2(22); /* This means this FUNCTION only supports up to 999999999 milliseconds, which is over 11 days */
d1_ NUMBER;
h1_ NUMBER;
m1_ NUMBER;
s1_ NUMBER;
ms_ NUMBER;
/* If you choose 1, you only get seconds. If you choose 2, you get minutes and seconds etc. */
precision_ NUMBER; /* 0 => milliseconds; 1 => seconds; 2 => minutes; 3 => hours; 4 => days */
format_ VARCHAR2(2) := ', ';
return_ VARCHAR2(50);
BEGIN
IF (syntax_ IS NULL) THEN
precision_ := 0;
ELSE
IF (syntax_ = 0) THEN
precision_ := 0;
ELSIF (syntax_ = 1) THEN
precision_ := 1;
ELSIF (syntax_ = 2) THEN
precision_ := 2;
ELSIF (syntax_ = 3) THEN
precision_ := 3;
ELSIF (syntax_ = 4) THEN
precision_ := 4;
ELSE
precision_ := 0;
END IF;
END IF;
SELECT EXTRACT(DAY FROM (end_ - start_)) INTO d1_ FROM DUAL;
SELECT EXTRACT(HOUR FROM (end_ - start_)) INTO h1_ FROM DUAL;
SELECT EXTRACT(MINUTE FROM (end_ - start_)) INTO m1_ FROM DUAL;
SELECT EXTRACT(SECOND FROM (end_ - start_)) INTO s1_ FROM DUAL;
IF (precision_ = 4) THEN
IF (d1_ = 1) THEN
day_ := ' day';
ELSE
day_ := ' days';
END IF;
IF (h1_ = 1) THEN
hour_ := ' hour';
ELSE
hour_ := ' hours';
END IF;
IF (m1_ = 1) THEN
minute_ := ' minute';
ELSE
minute_ := ' minutes';
END IF;
IF (s1_ = 1) THEN
second_ := ' second';
ELSE
second_ := ' seconds';
END IF;
return_ := d1_ || day_ || format_ || h1_ || hour_ || format_ || m1_ || minute_ || format_ || s1_ || second_;
RETURN return_;
ELSIF (precision_ = 3) THEN
h1_ := (d1_ * 24) + h1_;
IF (h1_ = 1) THEN
hour_ := ' hour';
ELSE
hour_ := ' hours';
END IF;
IF (m1_ = 1) THEN
minute_ := ' minute';
ELSE
minute_ := ' minutes';
END IF;
IF (s1_ = 1) THEN
second_ := ' second';
ELSE
second_ := ' seconds';
END IF;
return_ := h1_ || hour_ || format_ || m1_ || minute_ || format_ || s1_ || second_;
RETURN return_;
ELSIF (precision_ = 2) THEN
m1_ := (((d1_ * 24) + h1_) * 60) + m1_;
IF (m1_ = 1) THEN
minute_ := ' minute';
ELSE
minute_ := ' minutes';
END IF;
IF (s1_ = 1) THEN
second_ := ' second';
ELSE
second_ := ' seconds';
END IF;
return_ := m1_ || minute_ || format_ || s1_ || second_;
RETURN return_;
ELSIF (precision_ = 1) THEN
s1_ := (((((d1_ * 24) + h1_) * 60) + m1_) * 60) + s1_;
IF (s1_ = 1) THEN
second_ := ' second';
ELSE
second_ := ' seconds';
END IF;
return_ := s1_ || second_;
RETURN return_;
ELSE
ms_ := ((((((d1_ * 24) + h1_) * 60) + m1_) * 60) + s1_) * 1000;
IF (ms_ = 1) THEN
msecond_ := ' millisecond';
ELSE
msecond_ := ' milliseconds';
END IF;
return_ := ms_ || msecond_;
RETURN return_;
END IF;
END Core;
BEGIN
RETURN(Core(start_, end_, syntax_));
END Return_Elapsed_Time;
例如，如果我现在（12.10.2018 11:17:00.00）使用 Return_Elapsed_Time(TO_TIMESTAMP('12.04.2017 12:00:00.00', 'DD.MM.YYYY HH24:MI:SS .FF'),SYSTIMESTAMP)，它应该返回类似：
47344620000 milliseconds
【讨论】：
最好使用这样的程序：
CREATE OR REPLACE FUNCTION timestamp_diff
(
start_time_in TIMESTAMP
, end_time_in TIMESTAMP
)
RETURN NUMBER
AS
l_days NUMBER;
l_hours NUMBER;
l_minutes NUMBER;
l_seconds NUMBER;
l_milliseconds NUMBER;
BEGIN
SELECT extract(DAY FROM end_time_in-start_time_in)
, extract(HOUR FROM end_time_in-start_time_in)
, extract(MINUTE FROM end_time_in-start_time_in)
, extract(SECOND FROM end_time_in-start_time_in)
INTO l_days, l_hours, l_minutes, l_seconds
FROM dual;
l_milliseconds := l_seconds*1000 + l_minutes*60*1000 + l_hours*60*60*1000 + l_days*24*60*60*1000;
RETURN l_milliseconds;
END;
您可以通过调用来查看：
SELECT timestamp_diff (TO_TIMESTAMP('12.04.2017 12:00:00.00', 'DD.MM.YYYY HH24:MI:SS.FF'),
TO_TIMESTAMP('12.04.2017 12:00:01.111', 'DD.MM.YYYY HH24:MI:SS.FF'))
as milliseconds
FROM DUAL;
【讨论】：
这是一个存储过程：
CREATE OR REPLACE function timestamp_diff(a timestamp, b timestamp) return number is
begin
return extract (day from (a-b))*24*60*60 +
extract (hour from (a-b))*60*60+
extract (minute from (a-b))*60+
extract (second from (a-b));
end;
/
如果您还想击败否定他工作的 Oracle 开发人员的废话，请投票！
因为第一次比较时间戳每个人都需要一个小时左右...
【讨论】：
以上有语法错误，请在oracle上使用：
SELECT ROUND (totalSeconds / (24 * 60 * 60), 1) TotalTimeSpendIn_DAYS,
ROUND (totalSeconds / (60 * 60), 0) TotalTimeSpendIn_HOURS,
ROUND (totalSeconds / 60) TotalTimeSpendIn_MINUTES,
ROUND (totalSeconds) TotalTimeSpendIn_SECONDS
FROM
(SELECT ROUND ( EXTRACT (DAY FROM timeDiff) * 24 * 60 * 60 + EXTRACT (HOUR FROM timeDiff) * 60 * 60 + EXTRACT (MINUTE FROM timeDiff) * 60 + EXTRACT (SECOND FROM timeDiff)) totalSeconds
FROM
(SELECT TO_TIMESTAMP(TO_CHAR( date2 , 'yyyy-mm-dd HH24:mi:ss'), 'yyyy-mm-dd HH24:mi:ss') - TO_TIMESTAMP(TO_CHAR(date1, 'yyyy-mm-dd HH24:mi:ss'),'yyyy-mm-dd HH24:mi:ss') timeDiff
FROM TABLENAME
)
);
【讨论】：
在格式之间正确转换的时间戳，否则字段可能会被误解。
当从表 TableXYZ 中考虑两个不同的日期（Date2、Date1）时，这是一个正确的工作示例。
SELECT ROUND (totalSeconds / (24 * 60 * 60), 1) TotalTimeSpendIn_DAYS,
ROUND (totalSeconds / (60 * 60), 0) TotalTimeSpendIn_HOURS,
ROUND (totalSeconds / 60) TotalTimeSpendIn_MINUTES,
ROUND (totalSeconds) TotalTimeSpendIn_SECONDS
FROM (SELECT ROUND (
EXTRACT (DAY FROM timeDiff) * 24 * 60 * 60
+ EXTRACT (HOUR FROM timeDiff) * 60 * 60
+ EXTRACT (MINUTE FROM timeDiff) * 60
+ EXTRACT (SECOND FROM timeDiff))
totalSeconds,
FROM (SELECT TO_TIMESTAMP (
TO_CHAR (Date2,
'yyyy-mm-dd HH24:mi:ss')
- 'yyyy-mm-dd HH24:mi:ss'),
TO_TIMESTAMP (
TO_CHAR (Date1,
'yyyy-mm-dd HH24:mi:ss'),
'yyyy-mm-dd HH24:mi:ss')
timeDiff
FROM TableXYZ))
【讨论】：
SELECT numtodsinterval(date1-date2,'day') time_difference from dates;
SELECT (extract(DAY FROM time2-time1)*24*60*60)+
(extract(HOUR FROM time2-time1)*60*60)+
(extract(MINUTE FROM time2-time1)*60)+
extract(SECOND FROM time2-time1)
into diff FROM dual;
RETURN diff;
【讨论】：
当您减去两个TIMESTAMP 类型的变量时，您会得到一个INTERVAL DAY TO SECOND，其中包括取决于平台的毫秒数和/或微秒数。如果数据库在 Windows 上运行，systimestamp 通常会有毫秒。如果数据库在 Unix 上运行，systimestamp 通常会有微秒。
1 select systimestamp - to_timestamp( '2012-07-23', 'yyyy-mm-dd' )
2* from dual
SQL> /
SYSTIMESTAMP-TO_TIMESTAMP('2012-07-23','YYYY-MM-DD')
---------------------------------------------------------------------------
+000000000 14:51:04.339000000
您可以使用EXTRACT 函数提取INTERVAL DAY TO SECOND 的各个元素
SQL> ed
Wrote file afiedt.buf
1 select extract( day from diff ) days,
2 extract( hour from diff ) hours,
3 extract( minute from diff ) minutes,
4 extract( second from diff ) seconds
5 from (select systimestamp - to_timestamp( '2012-07-23', 'yyyy-mm-dd' ) diff
6* from dual)
SQL> /
DAYS HOURS MINUTES SECONDS
---------- ---------- ---------- ----------
0 14 55 37.936
然后您可以将这些组件中的每一个转换为毫秒并将它们相加
SQL> ed
Wrote file afiedt.buf
1 select extract( day from diff )*24*60*60*1000 +
2 extract( hour from diff )*60*60*1000 +
3 extract( minute from diff )*60*1000 +
4 round(extract( second from diff )*1000) total_milliseconds
5 from (select systimestamp - to_timestamp( '2012-07-23', 'yyyy-mm-dd' ) diff
6* from dual)
SQL> /
TOTAL_MILLISECONDS
------------------
53831842
不过，通常情况下，使用 INTERVAL DAY TO SECOND 表示或将小时、分钟、秒等分开的列比计算两个 TIMESTAMP 值之间的总毫秒数更有用。
【讨论】：
24*60*60，这是假设所有日子都是24H，这是错误的。在 PostgreSQL 中，我会减去从 EXTRACT(EPOCH FROM ...) 返回的两个值，但 Oracle 似乎没有等价物。
TIMESTAMP WITH TIME ZONE，则减去两者所产生的间隔应该考虑到任何时区/夏令时转换。因此，从太平洋时间凌晨 2 点的 TIMESTAMP WITH TIME ZONE 中减去东部时间凌晨 5 点的 TIMESTAMP WITH TIME ZONE 将得到 0 秒的间隔。如果您只是使用没有时区的标准TIMESTAMP，则没有时区作为数据的一部分存储，因此无法进行此类更正。
plsql 我试过这个：declare ta timestamp; select systimestamp - to_timestamp( '2012-07-23', 'yyyy-mm-dd' ) into ta from dual; 它返回Error(5,21): PL/SQL: ORA-00932: inconsistent datatypes: expected NUMBER got INTERVAL DAY TO SECOND
ta 需要是 interval day to second 数据类型，而不是 timestamp。从另一个中减去一个timestamp 返回一个interval 而不是另一个timestamp。