在我的 PHP 程序中,我$_SERVER
用来记录页面的访问日期:
$dateStamp = $_SERVER['REQUEST_TIME'];
结果是该$dateStamp
变量包含一个Unix 时间戳,如:
1385615749
将其转换为人类可读的日期/时间(带有年、月、日、小时、分钟、秒)的最简单方法是什么?
Your functional approch to convert timestamp into Human Readable format are as following
function convertDateTime($unixTime) {
$dt = new DateTime("@$unixTime");
return $dt->format('Y-m-d H:i:s');
}
$dateVarName = convertDateTime(1385615749);
echo $dateVarName;
Output :-
2013-11-28 05:15:49
<?php
$date = new DateTime();
$dateStamp = $_SERVER['REQUEST_TIME'];
$date->setTimestamp($dateStamp);
echo $date->format('U = Y-m-d H:i:s') . "\n";
?>
你可以试试这个
<?php
$date = date_create();
$dateStamp = $_SERVER['REQUEST_TIME'];
date_timestamp_set($date, $dateStamp);
echo date_format($date, 'U = D-M-Y H:i:s') . "\n";
?>
此代码将为您工作
$dateStamp = $_SERVER['REQUEST_TIME'];
echo date('d-M-Y H:i:s',strtotime($dateStamp));
REQUEST_TIME
- 它是 unix 时间戳 - 请求开始的时间戳。
$dateStamp = $_SERVER['REQUEST_TIME'];
echo date('d m Y', $dateStamp);
或者
$date = new DateTime($dateStamp);
echo $date->format('Y-m-d');