显示日期格式

Question

我希望展示2011年3月2日就像'02 -March-2011'.

Answer 1

<?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("j-F-Y");                          // 10-March-2001
    $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
    ?>

请参阅此链接....

http://php.net/manual/en/function.date.php

Answer 2

只需用您的值替换硬编码日期即可

$timestamp = strtotime('2-March-2011');
$newDate = date('d-F-Y', $timestamp); 
echo $newDate; //outputs 02-March-2011

Answer 3

您可以使用正则表达式、字符串分割sprintf()等，但我宁愿使用本机 DateTime 对象。

$str = '2-March-2011';

$date = DateTime::createFromFormat('j-F-Y', $str);

var_dump($date->format('d-F-Y')); // string(13) "02-March-2011" 

看见。