如何在PHP中将字符串转换为日期或日期时间

Question

my PHP script is receving "/Date(1403071826510)/" as a JSON date. How can I convert this to PHP Date or DateTime?

this is what I am doing:

$milliseconds = (int) preg_replace("/[^0-9 ]/", '', $strDate);
return json_encode(Date('d/m/Y h:m',$milliseconds/1000));

but my casting is returning 2147483647 instead of 1403071826510 because my OS is 32 bits.

Any idea how to get the Date or DateTime from the string received?

thanks

Answer 1

Strip off the milliseconds BEFORE you convert. Since it's a string, you just lop off the last 3 digits using substr() , and then cast to int

$js_timeval = '1403071826510';
$php_timeval = (int)substr($js_timeval, 0, -3); // 1403071826
echo date('r', $php_timeval);

Wed, 18 Jun 2014 00:10:26 -0600

Answer 2

There's really no need for a regex here. You can simply extract the seconds from the middle of the string and work with that:

$sec =substr("/Date(1403071826510)/", 6, 10);
$dt = date("Y-m-d H:i:s",$sec);                // 2014-06-18 06:10:26

Mindful of @zamnuts comment here's a regex version that handles early dates:

preg_match('#Date\(([0-9]{0,10}?)([0-9]{1,3}\))#',"/Date(1403071826510)/",$matches);
$dt = date("Y-m-d H:i:s", $matches[1]);

Answer 3

Marc B's answer works just fine (+1), but for completeness (and since I already did the work), I was thinking to use the BC Math extension for PHP , which must be installed and is not included by default. BC Math allows PHP to work with arbitrarily large numbers.

Consider the following:

$input = '/Date(1403071826510)/';

$ms = preg_replace('/[^0-9 ]/','',$input);
var_dump($ms); // string '1403071826510' (length=13)

$sec = bcdiv($ms,'1000');
var_dump($sec); // string '1403071826' (length=10)

$date = Date('d/m/Y h:m',$sec);
var_dump($date); // string '17/06/2014 11:06' (length=16)

All conversions use String . Divide the parsed milliseconds using bcdiv by '1000' and pass that into Date for the seconds timestamp.