PHP Date Format /Date(1365004652303-0500)/

PHP date format /Date(1365004652303-0500)/

First you need to understand the format you have

/Date(1365004652303-0500)/

Then you have

• time stamp (U) = 1365004652
• Milliseconds (u) = 303
• Difference to Greenwich time (GMT) (O) = -0500

Build a Format

$date = '/Date(1365004652303-0500)/';
preg_match('/(\d{10})(\d{3})([\+\-]\d{4})/', $date, $matches);
$dt = DateTime::createFromFormat("U.u.O",vsprintf('%2$s.%3$s.%4$s', $matches));
echo $dt->format('r');

Output

Wed, 03 Apr 2013 15:57:32 -0500
                            ^
                            |= Can you see the GMT ? 

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interface DateFormatParser
{
    /**
     * @param $string
     *
     * @return DateTime
     */
    public function parse($string);

}

abstract class PregDateParser implements DateFormatParser
{
    protected $pattern, $format, $mask;

    public function parse($string) {
        $string = (string)$string;

        $pattern = $this->pattern;
        $format  = $this->format;
        $mask    = $this->mask;

        $r = preg_match($pattern, $string, $matches);
        if (!$r) {
            throw new UnexpectedValueException('Preg Regex Pattern failed.');
        }
        $buffer = vsprintf($mask, $matches);
        $result = DateTime::createFromFormat($format, $buffer);
        if (!$result) {
            throw new UnexpectedValueException(sprintf('Failed To Create from Format "%s" for "%s".', $format, $buffer));
        }
        return $result;
    }
}

class JsonTimestampWithOffsetParser extends PregDateParser
{
    protected $pattern = '/^\/Date\((\d{10})(\d{3})([+-]\d{4})\)\/$/';
    protected $format  = 'U.u.O';
    protected $mask    = '%2$s.%3$s.%4$s';
}

$date   = '/Date(1365004652303-0500)/';
$parser = new JsonTimestampWithOffsetParser;
$dt     = $parser->parse($date);

echo $dt->format('r');

how to convert jsonecode /Date(1366585200000+0100)/ back to readable date

You could use this function:

function convertToDate($d, $format = 'm-d-Y') { 
    // First argument should have format like '/Date(1363824000000+0000)/'
    if (!preg_match("/[\d+-]+/", $d, $matches)) return null;
    return gmdate($format, $matches[0]/1000);
}

Call it like this:

print convertToDate('/Date(1363824000000+0000)/', 'm-d-Y');

Output:

03-21-2013

How to convert this API Date format into PHP date?

This is a timestamp in milisecondes, i suggest the snippet bellow to extract the date:

$text = 'Date(1549170000000-0500)';
preg_match('#\((.*?)\)#', $text, $match);

$mil = $match[1];
$seconds = $mil / 1000;
echo date("d/m/Y H:i:s", $seconds);

c#: Get timezone offset stored in date object

Simply use DateTimeOffset, not DateTime - it's a similar structure, only with an explicit Offset property. Plain DateTime doesn't have a concept of time zones.

This means using it within MyClass, and deserializing to it.

API returning invalid Date in PHP

Try this :

 var isWcfJsonDate = /\/Date(.*)\//.test("Your Date String");
        if (isWcfJsonDate)
        {
                dateStr = dateStr.match(/Date\((.*?)\)/)[1];
                dateValue = new Date(parseInt(dateStr));
                return dateValue;
        }
        dateValue = new Date(dateObject);
        console.log( dateValue); //Use your date value

Convert datetime c# to unixstamp in php

PHP's [strtotime][1] function is perfect for this. It can handle most common date formats, including strings.

Another option is with MySQL functions FROM_UNIXTIME and UNIX_TIMESTAMP

SELECT UNIX_TIMESTAMP(datetime_column) FROM table

This usually is faster than a PHP function call.

Parsing JSON DateTime from Newtonsoft's JSON Serializer

Use one of the JsonConverters that come with Json.NET for working with dates to get a better format. JavaScriptDateTimeConverter will automatically give you a JavaScript date.

public class LogEntry    
{    
  public string Details { get; set; }    
  public DateTime LogDate { get; set; }
}

[Test]
public void WriteJsonDates()
{    
  LogEntry entry = new LogEntry    
  {    
    LogDate = new DateTime(2009, 2, 15, 0, 0, 0, DateTimeKind.Utc),    
    Details = "Application started."    
  };    

  string defaultJson = JsonConvert.SerializeObject(entry);    
  // {"Details":"Application started.","LogDate":"\/Date(1234656000000)\/"}     

  string javascriptJson = JsonConvert.SerializeObject(entry, new JavaScriptDateTimeConverter());    
  // {"Details":"Application started.","LogDate":new Date(1234656000000)}

  string isoJson = JsonConvert.SerializeObject(entry, new IsoDateTimeConverter());    
  // {"Details":"Application started.","LogDate":"2009-02-15T00:00:00Z"}    
}

Documentation: Serializing Dates in JSON with Json.NET

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