On Sat, 30 Nov 2002, Randum Ian wrote: > Ministry is the name of the club. > Full is the type of the content. > 6122002 is the date - It is in the form ddmmyyyy so the example here is > 6th December 2002. > Is there a simple way I can just grab the date and sort it by order? > Should I change the format of the date to make it easier or something?
It'd be easier to sort if you changed your date format a bit ... what you have is rather difficult to work with, unless you take the string apart. (i.e. - How is '7102002' not greater than '6122002'?) For simplicity & ease, I work with date strings formatted as "YYYYMMDD". If your files were named "something-full-YYYYMMDD", you could get a sorted array of your files easily. $interestingFiles = array(); $dir = opendir( "/path/to/directory" ) or die( "Could not open dir" ); while( $dirEntry = readdir( $dir ) ){ if( ereg( "full-([0-9]{8})", $dirEntry, $MATCH ) ){ $interestingFiles[$MATCH[1]] = $dirEntry; } } ksort( $interestingFiles ); g.luck, ~Chris -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php