[SOLVED] Convert any date to yyyy-mm-dd format

[zohab]

Hi,

 

In my project user enter date in following format and i want to convert it into

yyyy-mm-dd format

 

 

  '[b]Y-m-d' => '2006-12-23',[/b]
    'm-d-Y' => '12-23-2006',
    'd-m-Y' => '23-12-2006',
    'Y/m/d' => '2006/12/23',
    'm/d/Y' => '12/23/2006',
    'd/m/Y' => '23/12/2006',
    'Y.m.d' => '2006.12.23',
    'd.m.Y' => '23.12.2006',
    'm.d.Y' => '12.23.2006',

 

examples

 

input=>24/08/2009

output=>2009-08-24

 

input=>08.24.2009

output=>2009-08-24

 

input=>24-08-2009

output=>2009-08-24

 

any function to perform above task

 

 

 

 

[5kyy8lu3]

i'm sleepy, but this is kinda what you'd need

 

<?php

function ConvertDate($OldDate)
{
$year = substr($OldDate, 6, 4);
$month = substr($OldDate, 0, 2);
$day = substr($OldDate, 2, 2);
$NewDate = $year . '-' . $month . '-' . $day;
return($NewDate);
}

echo ConvertDate('12/23/2006'); //this would output: 2006-12-23

//use a switch statement to choose the input format if you want to be able to convert different formats
//but basically you can use substr to pull each number off of the input string

[PravinS]

Use this

 

<?php
function format_date($date)
{
$dtarr = explode("/",$date);
return $dtarr[2]."-".$dtarr[1]."-".$dtarr[0];
}
echo format_date("24/08/2009");
?>

[zohab]

HI guys

 

echo ConvertDate('12/23/2006'); =>2006-12-/2

echo ConvertDate('2009/08/24'); =>8/24-20-09

 

not getting required result

[thebadbad]

The users also specify the format, right? Else it would be impossible to convert.

[zohab]

Hi

 

User do not specify format

 

we have to detect format and then convert

[5kyy8lu3]

Hi

 

User do not specify format

 

we have to detect format and then convert

that would be impossible...

 

how would you tell which is month and which is day for this date:

 

01/01/2009?  if the day is bigger than 12, you know it's the day since months only go to 12, but besides that, there's no way to tell unless the user specifies

[zohab]

Hi

 

OKay 

 

First I will check wheather we can get user entered date format and then convert

 

date to required format

 

 

 

 

[PFMaBiSmAd]

Several of the formats are ambiguous (you can interpret it more than one way) because you cannot tell from the value which is the month and which is the day.

 

You need to limit the format that can be entered by using three drop down select menus, one for the day, one for the month, and one for the year.

[Catfish]

or tell the user to input the date in your specific format (YYYY-MM-DD)

[PFMaBiSmAd]

You cannot rely on users to read, understand, and follow written instructions.

[zohab]

OKay

 

I got the format from user

 

examples

 

input=>m-d-Y

input=>08-24-2009

required output=>2009-08-24

 

input=>Y/m/d

input=>2009/08/24

required  output=>2009-08-24

 

 

input=>d.m.Y

input=>24-08-2009

required  output=>2009-08-24

 

any solution

[MadTechie]

The problem is this

 

given this date 07-03-08, what is the month ?

or making it easier 2007-03-08.. but still what's the month ?

 

the only option i can think of is

<?php
function ConvertDate($OldDate) {
return date("Y-m-d",strtotime($OldDate));
}
?>

which will return 2007-03-08 from my impossible example.

but 23.12.2006 would return 2006-12-23 so you MAY be fine..

 

[thebadbad]

In that case;

 

<?php
function standard_date($input, $format) {
$format = preg_replace('~[./-]~', '', $format);
$o = preg_split('~[./-]~', $input, 3);
switch($format) {
	case 'Ymd':
		return implode('-', $o);
	case 'mdY':
		return "$o[2]-$o[0]-$o[1]";
	case 'dmY':
		return "$o[2]-$o[1]-$o[0]";
	default:
		trigger_error('Wrong date format supplied to standard_date()', E_USER_ERROR);
}
}
//usage
echo standard_date('01.23.2008', 'm.d.Y');
?>

[zohab]

Hi,

 

Following not working

 

 

echo standard_date('24.08.2009', 'd.m.y');

echo standard_date('24/08/2009', 'd/m/y');

 

i am getting

 

Fatal error: Wrong date format supplied to standard_date()

[thebadbad]

That's because you're using lowercase Ys. Modified to be case insensitive:

 

<?php
function standard_date($input, $format) {
$format = strtolower(preg_replace('~[./-]~', '', $format));
$o = preg_split('~[./-]~', $input, 3);
switch($format) {
	case 'ymd':
		return implode('-', $o);
	case 'mdy':
		return "$o[2]-$o[0]-$o[1]";
	case 'dmy':
		return "$o[2]-$o[1]-$o[0]";
	default:
		trigger_error('Wrong date format supplied to standard_date()', E_USER_ERROR);
}
}
?>

[zohab]

Great work

 

thanks