来自/usr/share/postgresql/timezonesets/Default:
#################### AUSTRALIA ####################
ACSST 37800 D # Central Australia (not in zic)
AESST 39600 D # Australia Eastern Summer Standard Time (not in zic)
AEST 36000 # Australia Eastern Standard Time (not in zic)
AWSST 32400 D
我们有一个函数,它像JS一样在UTC中解析日期/时间对,但随后强制它的行为就像在本地时区中指定的那样。以下是:
var tz = (new Date()).toString().match(/\((.+)\)$/)[0];
var str = (new Date()).toUTCString();
str = str.replace(/GMT$/,tz);
var output = new Date(str);
正则表达式返回两个结果,对于我的时区,我得到'(Pacific Daylight Time)'和'Pacific Daylight Time'。对于我们
我正在试图找出最优的,并以最小的循环量的方式从这里分组我的js日期对象:(请注意,这是浏览器控制台输出,它是实际的JS日期,比如新日期())
[Sat Aug 08 2015 08:30:00 GMT+0200 (Central Europe Daylight Time), Sat Aug 08 2015 09:30:00 GMT+0200 (Central Europe Daylight Time), Sun Aug 09 2015 08:30:00 GMT+0200 (Central Europe Daylight Time), Sun Aug 09 2015 09:30:00 GMT+0
我正在存储一个日期集合,这些日期是从本月初开始的前12个月。所以我有:
my @t = localtime time();
my $m = $t[4];
my $y = $t[5];
foreach my $date (keys %$dates_ref) {
$m -= $comparison{$date}; # a hash of numbers to go back the correct number of months
$dates_ref->{$date} = mktime(0,0,0,1,$m,$y);
}
最后,我得到了一堆日期,如: Current mo
我试图改变背景和一个(1)走向不同的颜色后,在我的网页上一定时间(1700小时)。然而,它不会触发。我在这里做错什么了?
var today = new Date();
var time = today.getHours() + ":" + today.getMinutes() + ":" + today.getSeconds();
var dayLight = 0900;
var lightMode = 1700;
if (time >= dayLight && time <= lightMode) {
docume
假设,我有日期范围:
var start = moment('2017-08-21').startOf('isoweek').startOf('day');
var end = moment('2017-08-21').startOf('isoweek').add('days', 4).endOf('day');
我还有另一个数组:
var array = [
{x:Mon Aug 21 2017 08:00:00 GMT+0200 (Central European Dayl
我在Windows10专业版上使用的是带有clang编译器的c++ Builder10.2。有人能告诉我为什么这个不能编译吗?
// crt_tzset.cpp
// This program uses _tzset to set the global variables
// named _daylight, _timezone, and _tzname. Since TZ is
// not being explicitly set, it uses the system time.
#include <time.h>
#include <stdlib.h>
#i
为什么下列范围功能..。
d3.time.minute.range(
new Date('Sat Aug 17 2013 00:00:00 GMT-0400'),
new Date('Sat Aug 17 2013 06:00:00 GMT-0400'),
22);
...return像这样的数组..。
[
Sat Aug 17 2013 00:00:00 GMT-0400 (Eastern Daylight Time),
Sat Aug 17 2013 00:22:00 GMT-0400 (Eastern Daylight Time),
我的日期是2011-04-01格式的。当我运行下面的代码时,会得到以下结果。
var dateParts = '2011-04-01'.split('-');
dateParts[0] = parseInt(dateParts[0]);
dateParts[1] = parseInt(dateParts[1]);
dateParts[2] = parseInt(dateParts[2]);
console.log(dateParts);
for (i = 0; i <= 13; i++) {
var thisDate = new Date(datePa
var y = 0;
var a = [];
var ob = {};
for (var i = 0; i < rows.length - 1; i++) {
if (new Date(rows[i].date).getTime() === new Date(dates[y]).getTime()) {
ob.sum = parseInt(rows[i].calls);
ob.date = rows[i].date;
ob.time = rows[i].time;
a[y] = ob;
}
els