PHP - preg_match?

Question

I'm not very good with expressions... I've looked at some online tutorials, but I'm still not getting it. Basically, I'm trying to return TRUE if a string is formatted like this:

4 digits + space + 2 digits and convert it to a date.

So, the string will look like: 2010 02, and I'm trying to output February, 2010.

I'm trying to use preg_match, but I keep getting

{ is not a modifier...

EDIT

Per the first 2 responses, I changed it, but am getting a fatal error on the first and the same unknown modifier error on the second:

if (preg_match('/([0-9{4}]) ([0-9]{2})/iU', $path_part)) {
    $path_title = date("F, Y",strtotime(str_replace(" ", "-", $path_title)));
}

Also, just tried the more in-depth version in the first response, and while the error goes away, it doesn't change the output:

$path_part = '2010 02';
if (preg_match('/^(\d{4}) (\d{2})$/', $path_part, $matches)) {
    $path_title = $mon[(int)$matches[2]] . " " . $matches[1]; // prints Feb 2010
}

/([0-9{4}]) ([0-9]{2})/ is incorrect. Use /([0-9]{4}) ([0-9]{2})/ or /(\d{4}) (\d{2})/ — codaddict
– codaddict, Commented Feb 7, 2010 at 15:57

codaddict · Accepted Answer · 2010-02-07 15:51:34Z

3

I'm trying to return TRUE if a string is formatted like this: 4 digits + space + 2 digits

return preg_match(/^\d{4} \d{2}$/,$input);

To convert to date you can try something like:

$mon = array('','Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec');
$date_str = "2010 02";

if(preg_match('/^(\d{4}) (\d{2})$/',$date_str,$matches))
{
        print $mon[(int)$matches[2]] . " " . $matches[1]; // prints Feb 2010
}

edited Feb 7, 2010 at 15:51

answered Feb 7, 2010 at 15:40

codaddict

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2 Comments

phpN00b Over a year ago

Ok, I just did. I tried the longer explanation and the errors go away, but it doesn't output anything different. Still prints out 2010 02

Gumbo Over a year ago

You need to ensure that the number of the month is only in the valid range of 1..12.

bmb · Accepted Answer · 2010-02-07 16:28:34Z

1

$in = "2010 02";
if(preg_match('/([0-9]{4}) ([0-9]{2})/i', $in, $matches)) {
        echo date("F Y", strtotime($matches[2] . "/1/" . $matches[1]));
}

answered Feb 7, 2010 at 16:28

bmb

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Comments

Dan Soap · Accepted Answer · 2010-02-07 15:48:34Z

0

Try this one...

preg_match('/([0-9{4}]) ([0-9]{2})/iU', $input);

edited Feb 7, 2010 at 15:48

Dan Soap

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answered Feb 7, 2010 at 15:46

rATRIJS

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2 Comments

Dan Soap Over a year ago

This one's wrong. The first character group is mixed up with the mutliplier. It should be [0-9]{4} instead

phpN00b Over a year ago

I tried this, and I get the same error: warning: preg_match() [function.preg-match]: Unknown modifier '{'

Stephen Cross · Accepted Answer · 2010-02-07 15:52:05Z

0

Without having any detail as to your actual code, the following should work:

<?php

$str = '2010 02';

$months = array('January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December');

if(preg_match('/([0-9]{4}) ([0-9]{2})/', $str, $match) == 1){
    $year = $match[1];
    $month = (int) $match[2];
    echo $months[$month - 1] . ', ' . $year;
}else{
    //Error...
}

?>

answered Feb 7, 2010 at 15:52

Stephen Cross

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Comments

Danon · Accepted Answer · 2018-12-17 22:26:04Z

0

You can also use T-Regx library

$string = '2010 02';

pattern('\d{4} \d{2}')->match($string)->first(function (Match $match) 
{
    $year = $match->group(1);
    $month = $match->group(2);
});

answered Dec 17, 2018 at 22:26

Danon

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