(PHP 4, PHP 5, PHP 7)
getdate — 取得日期／时间信息
$timestamp = time()
] )
返回一个根据 timestamp
得出的包含有日期信息的关联数组 array。如果没有给出时间戳则认为是当前本地时间。
返回一个根据 timestamp
得出的包含有日期信息的关联数组 array。
返回的关联数组中的键名单元有以下几个：
| 键名 | 说明 | 返回值例子 |
|---|---|---|
| "seconds" | 秒的数字表示 | 0 到 59 |
| "minutes" | 分钟的数字表示 | 0 到 59 |
| "hours" | 小时的数字表示 | 0 到 23 |
| "mday" | 月份中第几天的数字表示 | 1 到 31 |
| "wday" | 星期中第几天的数字表示 | 0 (周日) 到 6 (周六) |
| "mon" | 月份的数字表示 | 1 到 12 |
| "year" | 4 位数字表示的完整年份 | 比如： 1999 或 2003 |
| "yday" | 一年中第几天的数字表示 | 0 到 365 |
| "weekday" | 星期几的完整文本表示 | Sunday 到 Saturday |
| "month" | 月份的完整文本表示，比如 January 或 March | January 到 December |
| 0 | 自从 Unix 纪元开始至今的秒数，和 time() 的返回值以及用于 date() 的值类似。 | 系统相关，典型值为从 -2147483648 到 2147483647。 |
Example #1 getdate() 例子
<?php
$today = getdate();
print_r($today);
?>
以上例程的输出类似于：
Array
(
[seconds] => 40
[minutes] => 58
[hours] => 21
[mday] => 17
[wday] => 2
[mon] => 6
[year] => 2003
[yday] => 167
[weekday] => Tuesday
[month] => June
[0] => 1055901520
)
Andre's code will throw an error. for the following line
$d = $todayh[mday];
$m = $todayh[mon];
$y = $todayh[year];
"Notice : Undefined constant mday ,mon,year"
As is, it was looking for constants called mday, mon, year etc. When it doesn't find such a constant, PHP interprets it as a string.
like any other request it should be wrapped in quotes like this
$d = $todayh['mday'];
$m = $todayh['mon'];
$y = $todayh['year'];
Here is another gmgetdate that is a little faster/suscint (no loops).
<?php
function gmgetdate2($ts = null){
$k = array('seconds','minutes','hours','mday',
'wday','mon','year','yday','weekday','month',0);
return(array_combine($k,split(":",
gmdate('s:i:G:j:w:n:Y:z:l:F:U',is_null($ts)?time():$ts))));
}
?>
It also returns the values in the same order as getdate.
It's worth noting that this is local time, not UTC/GMT - gmgetdate doesn't exist :(.
The most logical way to handle date arithmetic without hitting DST problems is to work in UTC...
<?php
function add_days($my_date,$numdays) {
$date_t = strtotime($my_date.' UTC');
return gmdate('Y-m-d',$date_t + ($numdays*86400));
}
?>
[it's even faster if you use gmmktime instead of strtotime]
I thought best to show a posseble way to go about bypassing the end month issue where the first day in a new month will have the monday of the week that it falls in - in the old month. Use the numbering of days as the constant and work you way from there.
Example:
<?php
//-----------------------------
$now = time();
$num = date("w");
if ($num == 0)
{ $sub = 6; }
else { $sub = ($num-1); }
$WeekMon = mktime(0, 0, 0, date("m", $now) , date("d", $now)-$sub, date("Y", $now)); //monday week begin calculation
$todayh = getdate($WeekMon); //monday week begin reconvert
$d = $todayh[mday];
$m = $todayh[mon];
$y = $todayh[year];
echo "$d-$m-$y"; //getdate converted day
?>
Allot less code makes everyone happy..
In addition to canby23 at ms19 post:
It's a very bad idea to consider day having 24 hours (86400 secs), because some days have 23, some - 25 hours due to daylight saving changes. Using of mkdate() and strtotime() is always preferred. strtotime() also has a very nice behaviour of datetime manipulations:
<?php
echo strtotime ("+1 day"), "\n";
echo strtotime ("+1 week"), "\n";
echo strtotime ("+1 week 2 days 4 hours 2 seconds"), "\n";
echo strtotime ("next Thursday"), "\n";
echo strtotime ("last Monday"), "\n";
?>