Question

我创建了一个脚本，通过该脚本我可以从rss提要中检索值。一个值采用以下日期格式：

Thu, 14 Apr 2011 18:23:19 GMT

我希望这个值符合以下格式：

April 14, 2011 11:53 PM

我怎样才能实现这个目标？

请帮我解决这个问题

由于

的Pankaj

Answer 1

DateTime课程非常方便。您可以使用其format()方法。

$datetime = new DateTime('Thu, 14 Apr 2011 18:23:19 GMT');

echo $datetime->format('F j,Y g:i A');

Answer 2

$olddate = "Thu, 14 Apr 2011 18:23:19 GMT";
$newdate = date("F d, Y H:i A",strtotime($oldDate));

Answer 3

试试这个，看看是否有帮助

echo(date('F j,Y g:i:A',strtotime(Date_From_RSS)));

Answer 4

试一试：

echo (date('F j,Y g:i:A',strtotime($Yourinputdate)));

Answer 5

<?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
?>

以上代码来自：http://www.php.net/manual/en/function.date.php

我没有看到你有输入所以这里是：

$date = new DateTime('Thu, 14 Apr 2011 18:23:19 GMT');
echo $date->format("F j, Y g:i a");

转换日期格式

5 个答案: