SELECT DATE_PART('day', '2011-12-31 01:00:00'::timestamp - '2011-12-29 23:00:00'::timestamp); -- Result: 1 1. 2. 3. 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: -- Difference between Dec 22, 2011 and Dec 31, 2011 in wee...
1、计算两个日期之间的天数差: SELECT date1 + INTERVAL '1 day' date2 AS days_diff; 2、计算两个日期之间的小时数差: SELECT (date1::time + INTERVAL '1 hour') (date2::time + INTERVAL '1 hour') AS hours_diff; 使用CAST和CONVERT函数计算时间差 我们需要将一个日期或时间值转换为另一个日期...
DATEDIFF ( datepart, {date|timestamp}, {date|timestamp} ) 周 selectdatediff(week,'2009-01-01','2009-12-31')asnumweeks; numweeks---52(1row) 季度 selectdatediff(qtr,'1998-07-01',current_date); date_diff---40(1row) PostgreSQL中时间和日期可以相互加减,得到同样的结果使用extract。 Postgre...
1. 查找两个日期之间的差异大于1天的记录 假设我们有一个表events,其中有两个日期字段start_date和end_date,我们可以使用date_diff函数来计算两个日期之间的差异。 代码语言:txt 复制 SELECT * FROM events WHERE date_diff('day', start_date, end_date) > 1; 2. 查找两个日期之间的差异...
可以看到,对于每一个用户而言,如果它的登陆日期sigindate是连续的,那么diff就会相同(具体是多少不重要)。 (2)对diff进行计数,即求出用户的所有连续活跃天数 select userid,diff, count(diff) as 'diff计数' from ( select *, datediff(now(), sigindate)-1267 as dates, ...
Date/time type storage: 64-bit integers Float4 argument passing: by value Float8 argument passing: by value Data page checksum version: 0 在standby查看wal接收情况 postgres=# \xon; Expanded displayison. postgres=#select*frompg_stat_wal_receiver;-[ RECORD1]---+---...
You are now connected to database"test"asuser"postgres".test=# select*from tb_mytps;i|name---+---1|name12|name23|name34|name45|name56|name67|name78|name89|name910|name10(10rows) 开始备份 建立备份文件存放路径 代码语言:javascript 复制 [root...
select extract(day FROM (age('2017-12-10'::date , '2017-12-01'::date))); 计算时间差秒数 select extract(epoch FROM (now() - (now()-interval '1 day') )); extract函数格式: extract (field from source) extract函数是从日期或者时间数值里面抽取子域,比如年、月、日等。source必须是timesta...
dateTemp timestamp; intervals interval; BEGIN IF lower($1) = lower(YEAR_CONST) THEN select cast(cast(incrementvalue as character varying) || ' year' as interval) into intervals; ELSEIF lower($1) = lower(MONTH_CONST) THEN select cast(cast(incrementvalue as character varying) || ' months...
SELECT (DATE'2001-02-16', DATE'2001-12-21') OVERLAPS (DATE'2001-10-30', DATE'2002-10-30'); 结果:trueSELECT(DATE'2001-02-16', INTERVAL'100 days')OVERLAPS (DATE'2001-10-30', DATE'2002-10-30'); 结果:falseSELECT(DATE'2001-10-29', DATE'2001-10-30')OVERLAPS ...