Converting dd/mm/yy hh:mm:ss into yyyy/mm/dd hh:mm:ss

[atzaman]

Hi

Please could someone kindly help me with an issue I'm having converting date formats.

I'd like to convert dates which are in the format mm/dd/yy hh:mm:ss into yyyy/mm/dd hh:mm:ss but am kind of stuck on how to do it.

Anyone able to help?

Many thanks in advance.

I suppose there are all sorts of nice conversion scripts out there. You can roll your own if you have a fixed input and output formats. A problem with your specific input format is the two year 'yy'. I assume that all dates need to be set to '20yy'?

You can split the input into components via substr():

function returnDate($input_date){
 //have a handling routine to check the validity of the input
 //a good way to do this would be with regular expressions - but I must admit, I struggle with these!
 //mm/dd/yy hh:mm:ss 
 //0123456789....... (character position - notice start at position 0)

 $month = substr($input_date,0,2);
 $day = substr($input_date,3,2);
 $year = '20' . substr($input_date,6,2);
 $time = substr($input_date,8);
 $output_date = $year . "/" . $month . "/" . $day . $time;
 return $output_date;
}

I haven't tested this! As mentioned, regular expressions could be of help here, cutting down the code to about two lines.

function returnDate($input_date){
//validate
 return $output_date = '20' . substr($input_date,6,2) . "/" . substr($input_date,0,2) . "/" . substr($input_date,3,2) . substr($input_date,8);
}

Of course, you could cut the the function to the above (one line), but that's just ugly!

[blocblue]

Hi,

How about trying the following?

function convertDate($strDate) {
    return date('Y/m/d G:i:s', strtotime($strDate));
}

R.

[atzaman]

I suppose there are all sorts of nice conversion scripts out there. You can roll your own if you have a fixed input and output formats. A problem with your specific input format is the two year 'yy'. I assume that all dates need to be set to '20yy'?

You can split the input into components via substr():

function returnDate($input_date){
 //have a handling routine to check the validity of the input
 //a good way to do this would be with regular expressions - but I must admit, I struggle with these!
 //mm/dd/yy hh:mm:ss 
 //0123456789....... (character position - notice start at position 0)

 $month = substr($input_date,0,2);
 $day = substr($input_date,3,2);
 $year = '20' . substr($input_date,6,2);
 $time = substr($input_date,8);
 $output_date = $year . "/" . $month . "/" . $day . $time;
 return $output_date;
}

I haven't tested this! As mentioned, regular expressions could be of help here, cutting down the code to about two lines.

function returnDate($input_date){
//validate
 return $output_date = '20' . substr($input_date,6,2) . "/" . substr($input_date,0,2) . "/" . substr($input_date,3,2) . substr($input_date,8);
}

Of course, you could cut the the function to the above (one line), but that's just ugly!

You, my friend, are awesome :)

Have to make a few changes to it but you've put me on the right track. Thank you!

[atzaman]

Hi,

How about trying the following?

function convertDate($strDate) {
    return date('Y/m/d G:i:s', strtotime($strDate));
}

R.

Many thanks for the reply. I'm pretty sure strotime only works when the date format is mm/dd/yy though...?

I agree with strtotime. It works with US format data - which is what you quoted in your first post and the format which I used in my solution. So technically, robothy is right - much nicer that my cobbled solution. However, the question in your title seems to be in UK/European format, so strtotime won't work with this.

As I mentioned, you could rework the function to work with all different kinds of format:

function($input_date,$input_format){
//$input_format could be 'us', 'euro' - depending on the parameter, you could do the following:
  if($input_format = 'us'){
    $output_date = date('Y/m/d G:i:s', strtotime($input_date));
  }else{
    $output_date = '20' . substr($input_date,6,2) . "/" . substr($input_date,3,2) . "/" . substr($input_date,0,2) . substr($input_date,8)
  }
  return $output_date;
}

BTW - made a mistake in my previous post:

function returnDate($input_date){
//validate
 return $output_date = '20' . substr($input_date,6,2) . "/" . substr($input_date,0,2) . "/" . substr($input_date,3,2) . substr($input_date,8);
}

should have been:

function returnDate($input_date){
//validate
 return '20' . substr($input_date,6,2) . "/" . substr($input_date,3,2) . "/" . substr($input_date,0,2) . substr($input_date,8);
}

No need for $output_date variable and swapped values to make UK/Euro format.

REM - if solved, mark it so. Cheers.

[atzaman]

I seem to have made an error in the actual description... should have stated dd/mm/yy like in the title of the thread. My error.

Yep, I'd already made that amendment to your code plus a couple of other changes :)

I thought I'd already marked this thread as solved. Is it still showing as unsolved?

Many thanks to you both once again and apologies to robothy as it was my mistake.

[developer!]

Thanks diafol :)

Im getting $dob in this format 26-DEC-85
I changed to 1985-12-26
I did like this:

/**
* parse the date of birth
*/

function person_details_dob( &$params , $dob)



    if (empty($dob))
    {
        return;
    }



    $day = substr($dob,0,2);
    $month = substr($dob,3,3);
    $yy = substr($dob,7,8);

    if($yy > 30){
        $year = '19' . $yy;
    }
    else
    {
        $year = '20' . $yy;
    }

    $months_array =  array(
                        'JAN'       => '01',
                        'FEB'       => '02',
                        'MAR'       => '03',
                        'APR'       => '04',
                        'MAY'       => '05',
                        'JUN'       => '06',
                        'JUL'       => '07',
                        'AUG'       => '08',
                        'SEP'       => '09',
                        'OCT'       => '10',
                        'NOV'       => '11',
                        'DEC'       => '12',
                        );
    if ($month)
    {
        $month = $months_array[$month];
    }

    return $dob = $year . "-" . $month . "-" . $day;

}

Im getting $dob in this format 26-DEC-85
I changed to 1985-12-26

Use this - it's easier

function formatDOB($date,$inputFormat='j-M-y',$outputFormat='Y-m-d')
{
    $d = date_create_from_format($inputFormat, $date);
    return date_format($d, $outputFormat);
}

echo formatDOB('26-DEC-85');

Obviously, you know to be careful with 2 digit years.

00-69 = 2000-2069
70-99 = 1970-1999

You could always set your own cutoff though.

Something like this:

function formatDOB($date,$yearcutoff=70,$inputFormat='j-M-y',$outputFormat='Y-m-d')
{
    $d = date_create_from_format($inputFormat, $date);
    if($yearcutoff != 70 && is_int($yearcutoff) && $yearcutoff < 100 && strpos($inputFormat,'y') !== false)
    {
        $y = date_format($d, 'y');
        if($y >= $yearcutoff && $y < 70){
            $d = date_sub($d, date_interval_create_from_date_string('100 years'));      
        }elseif($y < $yearcutoff && $y >= 70){
            $d = date_add($d, date_interval_create_from_date_string('100 years'));
        }
    }
    return date_format($d, $outputFormat);
}

//Usage

echo formatDOB('08-12-67',50,'d-m-y'); //gives: 1967-12-08

//without the yearcutoff it would give: 2067-12-08


/*TEST SCRIPT
for($i=0;$i<=99;$i++){
echo "<h3>Year CUTOFF: $i</h3>";

    for($x = 0;$x<=99;$x++)
    {
        $y = str_pad($x,2,'0',STR_PAD_LEFT);
        echo 'Shortyear:' . $y . " = " . formatDOB('26-Dec-' . $y, $i) . "<br />";
    }
}
*/

If you find a cutoff number that you're happy to stick with, it's easier to switch the parameter order:

function formatDOB($date,$inputFormat='j-M-y',$outputFormat='Y-m-d',$yearcutoff=40)