How to make strtotime parse dates in Australian (i.e. UK) format: dd/mm/yyyy?

问题

I can't beleive I've never come across this one before.

Basically, I'm parsing the text in human-created text documents and one of the fields I need to parse is a date and time. Because I'm in Australia, dates are formatted like dd/mm/yyyy but strtotime only wants to parse it as a US formatted date. Also, exploding by / isn't going to work because, as I mentioned, these documents are hand-typed and some of them take the form of d M yy.

I've tried multiple combinations of setlocale but no matter what I try, the language is always set to US English.

I'm fairly sure setlocale is the key here, but I don't seem to be able to strike upon the right code. Tried these:

au
au-en
en_AU
australia
aus

Anything else I can try?

For clarity: I'm running on IIS with a Windows box.

Thanks so much :)

Iain

Example:

$mydatetime = strtotime("9/02/10 2.00PM");
echo date('j F Y H:i', $mydatetime);

Produces

2 September 2010 14:00

I want it to produce:

9 February 2010 14:00

My solution

I'm giving the tick to one of the answers here as it is a much easier-to-read solution to mine, but here's what I've come up with:

$DateTime = "9/02/10 2.00PM";
$USDateTime = preg_replace('%([0-3]?[0-9]{1})\s*?[\./ ]\s*?((?:1[0-2])|0?[0-9])\s*?[./ ]\s*?(\d{4}|\d{2})%', '${2}/${1}/${3}', $DateTime);  
echo date('j F Y H:i',strtotime($USDateTime));

Because I can't rely on users to be consistent with their date entry, I've made my regex a bit more complex:

0 or 1 digit between 0 and 3
1 digit between 0 and 9 -- yes this will match 37 as a valid date but I think the regex is already big enough!
Could be some whitespace
Delimiting character (a '.', a '/' or a ' ')
Could be some whitespace
Either:
- A number between 10 and 12 OR
- A number between 1 and 9 with an optional leading 0
Could be some whitespace
Delimiting character (a '.', a '/' or a ' ')
Could be some whitespace
Either:
- A number 2 digits long OR
- A number 4 digits long

Hopefully this will match most styles of date writing...

回答1:

There's a quick fix that seems to force PHP's strtotime into using the UK date format. That is: to replace all of the '/' in the incoming string with a '-'.

Example:

date('Y-m-d', strtotime('01/04/2011'));

Would produce: 2011-01-04

date('Y-m-d', strtotime('01-04-2011'));

Would produce: 2011-04-01

You could use str_replace to achieve this.

Example:

str_replace('/', '-', '01/04/2011');

EDIT:

As stated in the comments, this works because PHP interprets slashes as American and dots/dashes as European.

I've used this trick extensively and had no problems so far.

If anyone has any good reasons not to use this, please comment.

回答2:

The problem is that strtotime doesn't take a format argument. What about strptime?

回答3:

setlocale() sucks for exactly the reason you describe: You never know what you're going to get. Best to process the string manually.

Zend Framework's Zend_Date is one alternative promising more exact and consistent date handling. I don't have experience with it myself yet, just beginning to work with it, but so far, I like it.

回答4:

Ah, the old problem us lucky Australians get.

What I've done in the past is something like this

public static function getTime($str) { // 3/12/2008

       preg_match_all('/^(\d{1,2})\/(\d{1,2})\/(\d{4})$/', $str, $matches);

       return (isset($matches[0][0])) ? strtotime($matches[3][0] . '-' . $matches[2][0] . '-' . $matches[1][0]) : NULL;

    }

Though this relies on dates in this format dd/mm/yyyy.

You can probably use another regex or so to convert from d M yy or use a modified one. I don't know if this would be correct but it may be a start:

/^(\d{1,2})(?:\/|\s)(\d{1,2})(?:\/|\s)(\d{2,4})$/

回答5:

$date = date_create_from_format('d/m/Y', $date_string);
$unix_timestamp = $date->getTimestamp();

$date = DateTime::createFromFormat('d/m/Y', $date_string);
$unix_timestamp = $date->getTimestamp();

回答6:

For future readers, try the following for parsing dates from unstrusted user input, for which strptime() is too rigid. It outputs ISO date strings which you can pass to new Date(), strtotime(), etc., or false for invalid dates, or null for no usable input. You'd have to adapt it to process times. It also accepts 2-digit year and ISO date input.

function parseUserDate($input, $allow_zero = false) {
    if (!is_string($input)) {
        return $input === null ? null : false;
    }
    $input = preg_split('/\\D+/', $input, null, PREG_SPLIT_NO_EMPTY);
    if (!$input) {
        return null;# no date
    }
    if (count($input) == 3) {
        list($od, , $oy) = $input;
        if (strlen($od) > 2 && strlen($oy) < 3) {# assume ISO
            $oy = $od;
            $input = array_reverse($input);
        }
        $intify = function($dig) {
            return (int) ltrim($dig, '0');# prevent octals
        };
        list($d, $m, $y) = array_map($intify, $input);
        if (strlen($oy) < 3) {
            $current_year = date('Y');
            $current_year_2 = $intify(substr($current_year, -2));
            $diff = $y - $current_year_2;
            if ($diff > 50) {
                $diff -= 100;
            }
            elseif ($diff < -50) {
                $diff += 100;
            }
            $y = $diff + $current_year;
        }
        if (
            ($allow_zero && $d == 0 && $m == 0 && $y == 0)
            || checkdate($m, $d, $y)# murrican function
        ) {
            return sprintf('%04d-%02d-%02d', $y, $m, $d);
        }
    }
    return false;# invalid date
}

来源：https://stackoverflow.com/questions/2444820/how-to-make-strtotime-parse-dates-in-australian-i-e-uk-format-dd-mm-yyyy

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strtotime