[SOLVED] grab name of file with most recent date in it

[jasonc]

I wish to get the file name which has the most recent date in it.

 

all files are in the following format... (dd/mm/yyy)

 

abc - 21-09-2009 - Agenda.pdf

 

but it is not getting the correct file for some reason, this is what code i am using to get the filename.

 

	$files = glob('agendas/*.pdf');
	$datestring = strtotime(basename(substr($files[0], 14, 10),".pdf"));
	$anchortext = date('j M Y',$datestring);
	// echo(".".basename(substr($files[0], 14, 10),".pdf").".");
	$url = date('dmY',$datestring);
		if (strlen(date('j',$datestring)) == 1) { ?>  <?
		} ?><strong><?=$anchortext;?></strong>

[MadTechie]

I would probably take this approach

$sortfile = array();
foreach($files as $file){
    //Extract date
    if (preg_match('/(\d+)-(\d+)-(\d+)[^.]*?\.pdf/', $file, $reg)) {
        //add date to array (file as key)
        $sortfile[$file] = mktime(0,0,0,$reg[2],$reg[1],$reg[3]);
    }
}
asort($sortfile); //Sort an array (the dates) and maintain index association (file names)

//echo files
foreach($sortfile as $file => $fdate){
   echo "$file<br>";
}

 

EDIT: updated regex and changed strtotime to mkdate

 

please note this is untested, but the last file should be the one you want, or use arsort and it will be the first one

 

 

[jasonc]

hey

 

thanks that sorted them in reverse order but how do i get the first file name in the array?

 

i tried $sortfile[0] and $sortfile[1]

 

but this did not work

[MadTechie]

your need to get the key, IE

$sortfile = array_keys($sortfile);
echo $sortfile[0];

[jasonc]

thats done it...

 

$sortfile = array_keys($sortfile);

echo $sortfile[0];

 

thanks for your help.

[jasonc]

this is how i made the post set to 'solved'

 

 

 

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