Convert a date format in PHP

Php Problem Overview

I am trying to convert a date from yyyy-mm-dd to dd-mm-yyyy (but not in SQL); however I don't know how the date function requires a timestamp, and I can't get a timestamp from this string.

How is this possible?

Php Solutions

Solution 1 - Php

Use strtotime() and date():

$originalDate = "2010-03-21";
$newDate = date("d-m-Y", strtotime($originalDate));

(See the strtotime and date documentation on the PHP site.)

Note that this was a quick solution to the original question. For more extensive conversions, you should really be using the DateTime class to parse and format :-)

Solution 2 - Php

If you'd like to avoid the strtotime conversion (for example, strtotime is not being able to parse your input) you can use,

$myDateTime = DateTime::createFromFormat('Y-m-d', $dateString);
$newDateString = $myDateTime->format('d-m-Y');

Or, equivalently:

$newDateString = date_format(date_create_from_format('Y-m-d', $dateString), 'd-m-Y');

You are first giving it the format $dateString is in. Then you are telling it the format you want $newDateString to be in.

Or if the source-format always is "Y-m-d" (yyyy-mm-dd), then just use DateTime:

<?php
    $source = '2012-07-31';
    $date = new DateTime($source);
    echo $date->format('d.m.Y'); // 31.07.2012
    echo $date->format('d-m-Y'); // 31-07-2012
?>

Solution 3 - Php

Use:

implode('-', array_reverse(explode('-', $date)));

Without the date conversion overhead, I am not sure it'll matter much.

Solution 4 - Php

$newDate = preg_replace("/(\d+)\D+(\d+)\D+(\d+)/","$3-$2-$1",$originalDate);

This code works for every date format.

You can change the order of replacement variables such $3-$1-$2 due to your old date format.

Solution 5 - Php

$timestamp = strtotime(your date variable); 
$new_date = date('d-m-Y', $timestamp);

For more, see the documentation for strtotime.

Or even shorter:

$new_date = date('d-m-Y', strtotime(your date variable));

Solution 6 - Php

Also another obscure possibility:

$oldDate = '2010-03-20'
$arr = explode('-', $oldDate);
$newDate = $arr[2].'-'.$arr[1].'-'.$arr[0];

I don't know if I would use it but still :)

Solution 7 - Php

There are two ways to implement this:

    $date = strtotime(date);

    $new_date = date('d-m-Y', $date);

    $cls_date = new DateTime($date);
    echo $cls_date->format('d-m-Y');

Solution 8 - Php

> Note: Because this post's answer sometimes gets upvoted, I came back > here to kindly ask people not to upvote it anymore. My answer is > ancient, not technically correct, and there are several better > approaches right here. I'm only keeping it here for historical > purposes. > > Although the documentation poorly describes the strtotime function, > @rjmunro correctly addressed the issue in his comment: it's in ISO > format date "YYYY-MM-DD". > > Also, even though my Date_Converter function might still work, I'd > like to warn that there may be imprecise statements below, so please > do disregard them.

~~The most voted answer is actually incorrect!~~

The PHP strtotime manual here states that "The function expects to be given a string containing an English date format". ~~What it actually means is that it expects an American US date format, such as "m-d-Y" or "m/d/Y".~~

That means that a date provided as "Y-m-d" may get misinterpreted by strtotime. You should provide the date in the expected format.

I wrote a little function to return dates in several formats. Use and modify at will. ~~If anyone does turn that into a class, I'd be glad if that would be shared.~~

function Date_Converter($date, $locale = "br") {

    # Exception
    if (is_null($date))
        $date = date("m/d/Y H:i:s");

    # Let's go ahead and get a string date in case we've
    # been given a Unix Time Stamp
    if ($locale == "unix")
        $date = date("m/d/Y H:i:s", $date);

    # Separate Date from Time
    $date = explode(" ", $date);

    if ($locale == "br") {
        # Separate d/m/Y from Date
        $date[0] = explode("/", $date[0]);
        # Rearrange Date into m/d/Y
        $date[0] = $date[0][1] . "/" . $date[0][0] . "/" . $date[0][2];
    }

    # Return date in all formats
        # US
        $Return["datetime"]["us"] = implode(" ", $date);
        $Return["date"]["us"]     = $date[0];

        # Universal
        $Return["time"]           = $date[1];
        $Return["unix_datetime"]  = strtotime($Return["datetime"]["us"]);
        $Return["unix_date"]      = strtotime($Return["date"]["us"]);
        $Return["getdate"]        = getdate($Return["unix_datetime"]);

        # BR
        $Return["datetime"]["br"] = date("d/m/Y H:i:s", $Return["unix_datetime"]);
        $Return["date"]["br"]     = date("d/m/Y", $Return["unix_date"]);

    # Return
    return $Return;

} # End Function

Solution 9 - Php

You can try the strftime() function. Simple example: strftime($time, '%d %m %Y');

Solution 10 - Php

Use this function to convert from any format to any format

function reformatDate($date, $from_format = 'd/m/Y', $to_format = 'Y-m-d') {
    $date_aux = date_create_from_format($from_format, $date);
    return date_format($date_aux,$to_format);
}

Solution 11 - Php

Given below is PHP code to generate tomorrow's date using mktime() and change its format to dd/mm/yyyy format and then print it using echo.

$tomorrow = mktime(0, 0, 0, date("m"), date("d") + 1, date("Y"));
echo date("d", $tomorrow) . "/" . date("m", $tomorrow). "/" . date("Y", $tomorrow);

Solution 12 - Php

date('m/d/Y h:i:s a',strtotime($val['EventDateTime']));

Solution 13 - Php

In PHP any date can be converted into the required date format using different scenarios for example to change any date format into Day, Date Month Year

$newdate = date("D, d M Y", strtotime($date));

It will show date in the following very well format

Mon, 16 Nov 2020

Solution 14 - Php

For this specific conversion we can also use a format string.

$new = vsprintf('%3$s-%2$s-%1$s', explode('-', $old));

Obviously this won't work for many other date format conversions, but since we're just rearranging substrings in this case, this is another possible way to do it.

Solution 15 - Php

Simple way Use strtotime() and date():

$original_dateTime = "2019-05-11 17:02:07"; #This may be database datetime
$newDate = date("d-m-Y", strtotime($original_dateTime));

With time

$newDate = date("d-m-Y h:i:s a", strtotime($original_dateTime));

Solution 16 - Php

You can change the format using the date() and the strtotime().

$date = '9/18/2019';

echo date('d-m-y',strtotime($date));

Result:

18-09-19

We can change the format by changing the ( d-m-y ).

Solution 17 - Php

Use date_create and date_format

Try this.

function formatDate($input, $output){
  $inputdate = date_create($input);
  $output = date_format($inputdate, $output);
  return $output;
}

Solution 18 - Php

function dateFormat($date)
{
    $m = preg_replace('/[^0-9]/', '', $date);
    if (preg_match_all('/\d{2}+/', $m, $r)) {
        $r = reset($r);
        if (count($r) == 4) {
            if ($r[2] <= 12 && $r[3] <= 31) return "$r[0]$r[1]-$r[2]-$r[3]"; // Y-m-d
            if ($r[0] <= 31 && $r[1] != 0 && $r[1] <= 12) return "$r[2]$r[3]-$r[1]-$r[0]"; // d-m-Y
            if ($r[0] <= 12 && $r[1] <= 31) return "$r[2]$r[3]-$r[0]-$r[1]"; // m-d-Y
            if ($r[2] <= 31 && $r[3] <= 12) return "$r[0]$r[1]-$r[3]-$r[2]"; //Y-m-d
        }

        $y = $r[2] >= 0 && $r[2] <= date('y') ? date('y') . $r[2] : (date('y') - 1) . $r[2];
        if ($r[0] <= 31 && $r[1] != 0 && $r[1] <= 12) return "$y-$r[1]-$r[0]"; // d-m-y
    }
}

var_dump(dateFormat('31/01/00')); // return 2000-01-31
var_dump(dateFormat('31/01/2000')); // return 2000-01-31
var_dump(dateFormat('01-31-2000')); // return 2000-01-31
var_dump(dateFormat('2000-31-01')); // return 2000-01-31
var_dump(dateFormat('20003101')); // return 2000-01-31

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