不同语言Unix时间戳和普通时间格式转化

如何实现Unix时间戳(Unix timestamp) 转化成普通时间格式？

Java String date = new java.text.SimpleDateFormat("dd/MM/yyyy HH:mm:ss").format(new java.util.Date(Unix timestamp * 1000))
JavaScript 先 var unixTimestamp = new Date(Unix timestamp * 1000) 然后 commonTime = unixTimestamp.toLocaleString()
Linux date -d @Unix timestamp
MySQL from_unixtime(Unix timestamp)
Perl 先 my $time = Unix timestamp 然后 my ($sec, $min, $hour, $day, $month, $year) = (localtime($time))[0,1,2,3,4,5,6]
PHP date('r', Unix timestamp)
PostgreSQL SELECT TIMESTAMP WITH TIME ZONE 'epoch' + Unix timestamp) * INTERVAL '1 second';
Python 先 import time 然后 time.gmtime(Unix timestamp)
Ruby Time.at(Unix timestamp)
SQL Server DATEADD(s, Unix timestamp, '1970-01-01 00:00:00')
VBScript / ASP DateAdd("s", Unix timestamp, "01/01/1970 00:00:00")
其他操作系统
(如果Perl被安装在系统中) 命令行状态：perl -e "print scalar(localtime(Unix timestamp))"

如何实现普通时间格式转化成 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)