Question

如何在不使用以下代码定义对象的情况下获得时间？

<?php
    $localtime = new DateTime("now", new DateTimeZone('Asia/Dhaka'));
    echo $localtime->format('H:i:s');
?>

Answer 1

尝试这种方法

以下是各种方法选择您的方法并尝试

<?php
// Assuming today is March 10th, 2001, 5:16:18 pm, and that we are in the
// Mountain Standard Time (MST) Time Zone

$today = date("F j, Y, g:i a");                 // March 10, 2001, 5:16 pm
$today = date("m.d.y");                         // 03.10.01
$today = date("j, n, Y");                       // 10, 3, 2001
$today = date("Ymd");                           // 20010310
$today = date('h-i-s, j-m-y, it is w Day');     // 05-16-18, 10-03-01, 1631 1618 6 Satpm01
$today = date('\i\t \i\s \t\h\e jS \d\a\y.');   // it is the 10th day.
$today = date("D M j G:i:s T Y");               // Sat Mar 10 17:16:18 MST 2001
$today = date('H:m:s \m \i\s\ \m\o\n\t\h');     // 17:03:18 m is month
$today = date("H:i:s");                         // 17:16:18
$today = date("Y-m-d H:i:s");                   // 2001-03-10 17:16:18 (the MySQL DATETIME format)
?>

Answer 2

像这样：

date_default_timezone_set('Asia/Dhaka');
echo date(' H:i:s');

如果您有权访问php.ini文件，那么您应该在那里设置时区。请记住重新启动服务器。

date.timezone = "Asia/Dhaka"

但使用DateTime类没有任何问题。

Answer 3

试试这个：

<?php
 $localtime = localtime();
 $localtime_assoc = localtime(time(), true);
 print_r($localtime);
 print_r($localtime_assoc);
?>

输出：

Array
(
    [0] => 24
    [1] => 3
    [2] => 19
    [3] => 3
    [4] => 3
    [5] => 105
    [6] => 0
    [7] => 92
    [8] => 1
)

Array
(
    [tm_sec] => 24
    [tm_min] => 3
    [tm_hour] => 19
    [tm_mday] => 3
    [tm_mon] => 3
    [tm_year] => 105
    [tm_wday] => 0
    [tm_yday] => 92
    [tm_isdst] => 1
)

关联数组的不同键的名称为如下：

[tm_sec] - seconds
[tm_min] - minutes
[tm_hour] - hour
[tm_mday] - day of the month
[tm_mon] - month of the year (January=0)
[tm_year] - Years since 1900
[tm_wday] - Day of the week (Sunday=0)
[tm_yday] - Day of the year
[tm_isdst] - Is daylight savings time in effect

看，如果有帮助。

获取PHP当地时间

3 个答案: