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linux下 如何将年月日时分秒格式转换为秒 - 99问答网

linux下 如何将年月日时分秒格式转换为秒

如题所述

第1个回答  2011-11-26
什么语言实现?
+ ====== 如何在不同编程语言中实现普通时间 → Unix时间戳(Unix timestamp)? ======
+
+ Java long epoch = new java.text.SimpleDateFormat("dd/MM/yyyy HH:mm:ss").parse("01/01/1970 01:00:00");
+
+ JavaScript var commonTime = new Date(Date.UTC(year, month - 1, day, hour, minute, second))
+
+ MySQL SELECT unix_timestamp(time) 时间格式: YYYY-MM-DD HH:MM:SS 或 YYMMDD 或 YYYYMMDD
+
+ Perl 先 use Time::Local 然后 my $time = timelocal($sec, $min, $hour, $day, $month, $year);
+
+ PHP mktime(hour, minute, second, day, month, year)
+
+ PostgreSQL SELECT extract(epoch FROM date('YYYY-MM-DD HH:MM:SS'));
+
+ Python 先 import time 然后 int(time.mktime(time.strptime('YYYY-MM-DD HH:MM:SS', '%Y-%m-%d %H:%M:%S')))
+
+ Ruby Time.local(year, month, day, hour, minute, second)
+
+ SQL Server SELECT DATEDIFF(s, '1970-01-01 00:00:00', time)
+
+ Unix / Linux date +%s -d"Jan 1, 1970 00:00:01"
+
+ VBScript / ASP DateDiff("s", "01/01/1970 00:00:00", time)追问

shell 呢?

追答

上面有啊 基本上所有语言都包括了

第2个回答  2015-08-04
date +%s -d'2015-07-01 00:00:00'
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