識別日期格式

Question

這是什麼日期格式？

2012-06-08dT00:00:00Z 

我該如何將php中的時間戳轉換爲此日期格式？

Answer 1

這裏是東西給你，從開始：

$date = "2012-06-08dT01:02:03Z"; 

// parse the date correctly 
$parsed_date = date_parse_from_format('Y-m-d H:i:s ', $date); 
print_r($parsed_date); 

// Make time from parsed date 
$old_date = mktime($parsed_date['hour'], $parsed_date['minute'], $parsed_date['second'], $parsed_date['month'], $parsed_date['day'], $parsed_date['year']); 

$now_date = time(); 

// a silly way to print that parsed date into the original way as before 
echo date("Y-m-d", $old_date) . 'dT' . date("H:i:s", $old_date) . 'Z'; 
echo "\n"; 

// a silly way to print current date/time in that format 
echo date("Y-m-d", $now_date) . 'dT' . date("H:i:s", $now_date) . 'Z'; 

Answer 2

$dt = new DateTime('2012-06-08T00:00:00Z'); //with no 'd' 
$timestamp = $dt->format('U'); 

如果你必須有 'd'，那麼：

$dt = DateTime::createFromFormat('Y-m-d??H:i:s?', '2012-06-08dT00:00:00Z'); 
$timestamp = $dt->format('U'); 

ETA：時間戳 - >您的格式

$dt = new DateTime('@1339124400'); //the @ indicates the following number is a timestamp 
$isoformat= $dt->format('Y-m-d\TH:i:sZ'); //leave out the 'd' and escape the 'T'