PostgreSQL-DATEDIFF-日期时间差,以秒,天,月,周等为单位
您可以使用各种日期时间表达式或用户定义的 DATEDIFF 函数(UDF)在 PostgreSQL 中计算两个日期时间值之间的差,以秒,分钟,小时,天,周,月和年为单位。
总览
PostgreSQL 不提供类似于 SQL Server DATEDIFF 的[2] DATEDIFF 函数,但是您可以使用各种表达式或 UDF 来获得相同的结果。
SQL Server and Sybase |
PostgreSQL | |
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Years | DATEDIFF(yy, start, end) | DATE_PART('year', end) - DATE_PART('year', start) |
Months | DATEDIFF(mm, start, end) | years_diff * 12 + (DATE_PART('month', end) - DATE_PART('month', start)) |
Days | DATEDIFF(dd, start, end) | DATE_PART('day', end - start) |
Weeks | DATEDIFF(wk, start, end) | TRUNC(DATE_PART('day', end - start)/7) |
Hours | DATEDIFF(hh, start, end) | days_diff * 24 + DATE_PART('hour', end - start ) |
Minutes | DATEDIFF(mi, start, end) | hours_diff * 60 + DATE_PART('minute', end - start ) |
Seconds | DATEDIFF(ss, start, end) | minutes_diff * 60 + DATE_PART('minute', end - start ) |
PostgreSQL-年中的日期差异
考虑使用 SQL Server 函数来计算以年为单位的两个日期之间的差:
SQL Server:
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-- Difference between Oct 02, 2011 and Jan 01, 2012 in years SELECT DATEDIFF( year , '2011-10-02' , '2012-01-01' ); -- Result: 1 |
请注意,SQL Server DATEDIFF 函数返回 1 年,尽管日期之间只有 3 个月。
SQL Server 不计算日期之间经过的整年,它仅计算年份之间的差异。
在 PostgreSQL 中,您可以从日期中获取年份部分并将其减去。
PostgreSQL:
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-- Difference between Oct 02, 2011 and Jan 01, 2012 in years SELECT DATE_PART( 'year' , '2012-01-01' :: date ) - DATE_PART( 'year' , '2011-10-02' :: date ); -- Result: 1 |
PostgreSQL-月中的日期差异
考虑使用 SQL Server 函数来计算两个日期(以月为单位)之间的差额:
SQL Server:
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-- Difference between Oct 02, 2011 and Jan 01, 2012 in months SELECT DATEDIFF( month , '2011-10-02' , '2012-01-01' ); -- Result: 3 |
在 PostgreSQL 中,您可以将年份之间的差值乘以 12,然后将月份部分之间的差值相加(可以为负)。
PostgreSQL:
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-- Difference between Oct 02, 2011 and Jan 01, 2012 in months SELECT (DATE_PART( 'year' , '2012-01-01' :: date ) - DATE_PART( 'year' , '2011-10-02' :: date )) * 12 + (DATE_PART( 'month' , '2012-01-01' :: date ) - DATE_PART( 'month' , '2011-10-02' :: date )); -- Result: 3 |
PostgreSQL-日期的天数差异
考虑使用 SQL Server 函数来计算两天之间的日期差:
SQL Server:
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-- Difference between Dec 29, 2011 23:00 and Dec 31, 2011 01:00 in days SELECT DATEDIFF( day , '2011-12-29 23:00:00' , '2011-12-31 01:00:00' ); -- Result: 2 |
请注意,DATEDIFF 返回了 2 天,尽管 datetime 值之间只有 1 天 2 小时。
在 PostgreSQL 中,如果您从另一个中减去一个日期时间值(TIMESTAMP,DATE 或 TIME 数据类型),则将获得一个 INTERVAL 值,格式为“ ddd days hh:mi:ss ”。
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SELECT '2011-12-31 01:00:00' :: timestamp - '2011-12-29 23:00:00' :: timestamp ; -- Result: "1 day 02:00:00" SELECT '2011-12-31 01:00:00' :: timestamp - '2010-09-17 23:00:00' :: timestamp ; -- Result: "469 days 02:00:00" |
所以,你可以使用 date_part 数函数 extact 的天数,但它返回的数量充分的日期之间的天数。
PostgreSQL:
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-- Difference between Dec 29, 2011 23:00 and Dec 31, 2011 01:00 in days SELECT DATE_PART( 'day' , '2011-12-31 01:00:00' :: timestamp - '2011-12-29 23:00:00' :: timestamp ); -- Result: 1 |
PostgreSQL-周中的日期差异
考虑使用 SQL Server 函数来计算两周中两个日期之间的差额:
SQL Server:
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-- Difference between Dec 22, 2011 and Dec 31, 2011 in weeks SELECT DATEDIFF(week, '2011-12-22' , '2011-12-31' ); -- Result: 1 |
DATEDIFF 返回日期时间值之间的整周数。
在 PostgreSQL 中,您可以使用表达式定义天数(请参见上文)并将其除以 7。需要 TRUNC 才能删除除后的小数部分。
PostgreSQL:
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-- Difference between Dec 22, 2011 and Dec 31, 2011 in weeks SELECT TRUNC(DATE_PART( 'day' , '2011-12-31' :: timestamp - '2011-12-22' :: timestamp )/7); -- Result: 1 |
PostgreSQL-日期时间的小时差异
考虑使用 SQL Server 函数来计算两个 datetime 值之间的时差,以小时为单位:
SQL Server:
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-- Difference between Dec 30, 2011 08:55 and Dec 30, 2011 9:05 in weeks SELECT DATEDIFF( hour , '2011-12-30 08:55' , '2011-12-30 09:05' ); -- Result: 1 |
请注意,尽管 datetime 值之间只有 10 分钟的差异,但 DATEDIFF 返回了 1 小时。
在 PostgreSQL 中,您可以使用表达式来定义天数(请参见上文),乘以 24 并乘以小时。
PostgreSQL:
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-- Difference between Dec 30, 2011 08:55 and Dec 30, 2011 9:05 in weeks SELECT DATE_PART( 'day' , '2011-12-30 08:55' :: timestamp - '2011-12-30 09:05' :: timestamp ) * 24 + DATE_PART( 'hour' , '2011-12-30 08:55' :: timestamp - '2011-12-30 09:05' :: timestamp ); -- Result: 0 |
请注意,此 PostreSQL 表达式返回在 datetime 值之间传递的完整小时数。
PostgreSQL-分钟中的日期时间差异
考虑使用 SQL Server 函数以分钟为单位计算两个日期时间值之间的差:
SQL Server:
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-- Difference between Dec 30, 2011 08:54:55 and Dec 30, 2011 08:56:10 in minutes SELECT DATEDIFF( minute , '2011-12-30 08:54:55' , '2011-12-30 08:56:10' ); -- Result: 2 -- Time only SELECT DATEDIFF( minute , '08:54:55' , '08:56:10' ); -- Result: 2 |
请注意,尽管 datetime 值之间只有 1 分 15 秒,但 DATEDIFF 返回了 2 分钟。
在 PostgreSQL 中,您可以使用一个表达式来定义小时数(请参阅上文),乘以 60 并乘以分钟。
PostgreSQL:
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-- Difference between Dec 30, 2011 08:54:55 and Dec 30, 2011 08:56:10 in minutes SELECT (DATE_PART( 'day' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp ) * 24 + DATE_PART( 'hour' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp )) * 60 + DATE_PART( 'minute' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp ); -- Result: 1 -- Time only SELECT DATE_PART( 'hour' , '08:56:10' :: time - '08:54:55' :: time ) * 60 + DATE_PART( 'minute' , '08:56:10' :: time - '08:54:55' :: time ); -- Result: 1 |
请注意,这些 PostreSQL 表达式返回在 datetime 值之间传递的完整分钟数。
PostgreSQL-日期时间差(以秒为单位)
考虑使用 SQL Server 函数以秒为单位计算两个日期时间值之间的差:
SQL Server:
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-- Difference between Dec 30, 2011 08:54:55 and Dec 30, 2011 08:56:10 in seconds SELECT DATEDIFF( second , '2011-12-30 08:54:55' , '2011-12-30 08:56:10' ); -- Result: 75 -- Time only SELECT DATEDIFF( second , '08:54:55' , '08:56:10' ); -- Result: 75 |
在 PostgreSQL 中,您可以使用表达式定义分钟数(请参见上文),乘以 60 并乘以秒。
PostgreSQL:
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-- Difference between Dec 30, 2011 08:54:55 and Dec 30, 2011 08:56:10 in seconds SELECT ((DATE_PART( 'day' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp ) * 24 + DATE_PART( 'hour' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp )) * 60 + DATE_PART( 'minute' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp )) * 60 + DATE_PART( 'second' , '2011-12-30 08:56:10' :: timestamp - '2011-12-30 08:54:55' :: timestamp ); -- Result: 75 -- Time only SELECT (DATE_PART( 'hour' , '08:56:10' :: time - '08:54:55' :: time ) * 60 + DATE_PART( 'minute' , '08:56:10' :: time - '08:54:55' :: time )) * 60 + DATE_PART( 'second' , '08:56:10' :: time - '08:54:55' :: time ); -- Result: 75 |
PostgreSQL DATEDIFF-用户定义函数(UDF)
除了使用单独的表达式来计算每个时间单位的日期时间差之外,还可以使用类似于 SQL Server DATEDIFF 函数的函数。
PostgreSQL:
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CREATE OR REPLACE FUNCTION DateDiff (units VARCHAR (30), start_t TIMESTAMP , end_t TIMESTAMP ) RETURNS INT AS $$ DECLARE diff_interval INTERVAL; diff INT = 0; years_diff INT = 0; BEGIN IF units IN ( 'yy' , 'yyyy' , 'year' , 'mm' , 'm' , 'month' ) THEN years_diff = DATE_PART( 'year' , end_t) - DATE_PART( 'year' , start_t); IF units IN ( 'yy' , 'yyyy' , 'year' ) THEN -- SQL Server does not count full years passed (only difference between year parts) RETURN years_diff; ELSE -- If end month is less than start month it will subtracted RETURN years_diff * 12 + (DATE_PART( 'month' , end_t) - DATE_PART( 'month' , start_t)); END IF; END IF; -- Minus operator returns interval 'DDD days HH:MI:SS' diff_interval = end_t - start_t; diff = diff + DATE_PART( 'day' , diff_interval); IF units IN ( 'wk' , 'ww' , 'week' ) THEN diff = diff/7; RETURN diff; END IF; IF units IN ( 'dd' , 'd' , 'day' ) THEN RETURN diff; END IF; diff = diff * 24 + DATE_PART( 'hour' , diff_interval); IF units IN ( 'hh' , 'hour' ) THEN RETURN diff; END IF; diff = diff * 60 + DATE_PART( 'minute' , diff_interval); IF units IN ( 'mi' , 'n' , 'minute' ) THEN RETURN diff; END IF; diff = diff * 60 + DATE_PART( 'second' , diff_interval); RETURN diff; END ; $$ LANGUAGE plpgsql; |
如何使用 PostgreSQL DATEDIFF 函数
语法与 SQL Server DATEDIFF 相似,但是您必须在 PostgreSQL 中将时间单位(秒,分钟等及其缩写)指定为字符串文字,例如:
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-- Difference between Dec 30, 2011 08:54:55 and Dec 30, 2011 08:56:10 in seconds SELECT DATEDIFF( 'second' , '2011-12-30 08:54:55' :: timestamp , '2011-12-30 08:56:10' :: timestamp ); -- Result: 75 |
PostgreSQL DATEDIFF 函数仅适用于 TIME
您可以具有另一个仅对时间数据类型起作用的函数。PostgreSQL 支持具有相同名称但参数数据类型不同的重载函数:
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CREATE OR REPLACE FUNCTION DateDiff (units VARCHAR (30), start_t TIME , end_t TIME ) RETURNS INT AS $$ DECLARE diff_interval INTERVAL; diff INT = 0; BEGIN -- Minus operator for TIME returns interval 'HH:MI:SS' diff_interval = end_t - start_t; diff = DATE_PART( 'hour' , diff_interval); IF units IN ( 'hh' , 'hour' ) THEN RETURN diff; END IF; diff = diff * 60 + DATE_PART( 'minute' , diff_interval); IF units IN ( 'mi' , 'n' , 'minute' ) THEN RETURN diff; END IF; diff = diff * 60 + DATE_PART( 'second' , diff_interval); RETURN diff; END ; $$ LANGUAGE plpgsql; |
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例如,可以将此函数调用为:
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-- Difference between 08:54:55 and 08:56:10 in seconds SELECT DATEDIFF( 'second' , '08:54:55' :: time , '08:56:10' :: time ); -- Result: 75 |
总结
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原文链接:https://blog.csdn.net/qq_35222843/article/details/127927404