preg_match php, forming the pattern

Question

So out of all of my php skills, I just suck at regular expressions. I totally understand the concept of it, but I can't write my patterns well. Well I need them for to extract date information from a file path. Here's my code:

#example path   
$path='/Users/landonsilla/Desktop/HS Pilot Project/Archive/Cabinet/Documents/2006-11In/Car Park/AR190298.doc';

echo $path."\n\n";
$pattern1 = '/[0-9]{4}[0-9]{2}[0-9]{2}/';#match YYYYMMDD
$pattern2 = '/[0-9]{4} [0-9]{2} [0-9]{2}/';#match YYYY MM DD
$pattern3 = '/[0-9]{4}[-][0-9]{2}/';#match YYYY-MM
$pattern4 = '/20[0-9]{2}/';#match YYYY

if(preg_match($pattern1, $path, $matches)){
    $ret=$matches[0].'-'.$matches[1].'-'.$matches[2].' 00:00:00';
}
else if(preg_match($pattern2, $path, $matches)){
    $ret=$matches[0].'-'.$matches[1].'-'.$matches[2].' 00:00:00';
}
else if(preg_match($pattern3, $path, $matches)){
    $ret=$matches[0].'-'.$matches[1].'-01 00:00:00';
}
else if(preg_match($pattern4, $path, $matches)){
    $ret=$matches[0].'-01-01 00:00:00';
}
else{
    $ret=date ("Y-m-d H:i:s.", filemtime($path));
}

The $pattern1,2,3,4 variables just don't work. Can somebody please let me know how to change those variables such that I can wind up having $ret be a date in my standard (mysql) format?

You have no groups for the suppatterns of which matches you use. var_dump() is your friend, start with single patterns in a test-file, debug until you get your results. — hakre
– hakre, Commented Oct 10, 2011 at 23:07
well i'm using print_r() (same thing as var_dump in this case), but i can't seem to get multiple values out of the one string with one function call — Landon
– Landon, Commented Oct 10, 2011 at 23:10
@deceze, thanks for the url, that's a fun tool to play with :) — Landon
– Landon, Commented Oct 10, 2011 at 23:22

Marcus · Accepted Answer · 2011-10-10 23:12:52Z

1

First off you need to group your subpatterns using (). This will store the results in the matches[] array as you assumed. Further more the matches are stored in index 1 and up so you need to add 1 to each of your matches in your if/else.

$path='/Users/landonsilla/Desktop/HS Pilot Project/Archive/Cabinet/Documents/2002-11In/Car Park/AR190298.doc';

echo $path."\n\n";
$pattern1 = '/([0-9]{4})([0-9]{2})([0-9]{2})/';#match YYYYMMDD
$pattern2 = '/([0-9]{4})([0-9]{2})([0-9]{2})/';#match YYYY MM DD
$pattern3 = '/([0-9]{4})[-]([0-9]{2})/';#match YYYY-MM
$pattern4 = '/(20[0-9]{2})/';#match YYYY

if(preg_match($pattern1, $path, $matches)){
    $ret=$matches[1].'-'.$matches[2].'-'.$matches[3].' 00:00:00';
}
else if(preg_match($pattern2, $path, $matches)){
    $ret=$matches[1].'-'.$matches[2].'-'.$matches[3].' 00:00:00';
}
else if(preg_match($pattern3, $path, $matches)){
    $ret=$matches[1].'*-*'.$matches[2].'-01 00:00:00';
}
else if(preg_match($pattern4, $path, $matches)){
    $ret=$matches[1].'-01-01 00:00:00';
}
else{
    $ret=date ("Y-m-d H:i:s.", filemtime($path));
}

Result can be viewed on here.

answered Oct 10, 2011 at 23:12

Marcus

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1 Comment

Landon Over a year ago

great! that's exactly what i wanted, i looks like i was close, i just didn't know about the grouping with parens and the additional offset in the resulting array.

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