php script that populates photos to a html div

Question

I currently have a picture page with about 250 jpeg. Images and I currently have a web page where I manually assigned a picture to a section on the page, however this is very ti

Answer 1

<?php

$test = '1234_1_01.jpeg';
list($pict_id, $quad_id) = explode('_', $test);
echo "Pict ID: $pict_id<br />
      Quad ID: $quad_id";

Answer 2

Your best bet is to go through the whole folder and filter the pictures accordingly like so:

$myPicFolder = "path/to/pictures/";
$thePics = array();
if (is_dir($myPicFolder)) {
    if ($dh = opendir($myPicFolder)) {
        while (($file = readdir($dh)) !== false)) {
            if(!is_dir($myPicFolder . $file){
                list($intersectionID,$section,$imgID) = explode('_'.$file);//Seperates the parts of the name for easier access
                //check if the ID is already a key in $thePics
                if(!array_key_exists($intersectionID,$thePics)){
                    //Add the new ID if it does not exist with an empty array.
                    $thePics[$intersectionID]=array();
                }
                if(!array_key_exists($section,$thePics[$intersectionID])){
                    //Add the new ID if it does not exist with an empty array.
                    $thePics[$intersectionID][$section]=array();
                }
                // Add a new part to the ID since each picture ID has 4 pictures.
                $thePics[$intersectionID][$section][] = $file;
            }
        }
        closedir($dh);
    }
}


//$thePics will then look like:
// array('1234'=>array(
//        '1'=>'1234_1_01.jpeg',
//        '2'=>'1234_2_03.jpeg',
//        '3'=>'1234_3_02.jpeg',
//        '4'=>'1234_4_01.jpeg'),
//    '345422'=>array(
//        '1'=>'345422_1_01.jpeg',
//        '2'=>'345422_2_02.jpeg')
//    );

//Cycle through the pictures...
foreach($thePics as $pictureID=>$quad){
    echo '<div class="intersection">';
    foreach($quad as $quadPart=>$pictures){
        echo '<div class=”s1”>';
        foreach($pictures as $pictureSRC){
            echo '<img src="'.$pictureSRC.'"/>';
        }
        echo '</div>';
    }
    echo '</div>';
}

Please remember to do some more verifications on the file natures to make sure you are only sending pictures... Maybe test on file extension and mime type...

EDIT: Added some comments and used list() to give explicit names to the variables I am using for better understanding.