Display all images from an external root folder using PHP

Question

I want to display all the images that are stored outside my web root folder. Please help me. I can only display one image. For example, if I have 5 images in my folder, only one image is displayed 5 times in my browser. Please help me with this. I have been working on this issue for over a month now. I'm a newbie. Help. Thank. Here is the code I'm using.

images.php

<?php   
  // Get our database connector
require("includes/copta.php");

// Grab the data from our people table
$sql = "select * from people";

$result = mysql_query($sql) or die ("Could not access DB: " . mysql_error());

$imgLocation = " /uploadfile/";

while ($row = mysql_fetch_array($result))
{
    $imgName = $row["filename"]; 
    $imgPath = $imgLocation . $imgName;

    echo "<img src=\"call_images.php?imgPath=" . $imgName . "\"  alt=\"\"><br/>";
    echo $row['id'] . " " . $imgName. "<br />";

}

?>

call_images.php

<?php
  // Get our database connector
require("includes/copta.php");

$imgLocation = '/ uploadz/';

$sql = "select * from people";

$result = mysql_query($sql) or 
    die ("Could not access DB: " . mysql_error());   

while ($row = mysql_fetch_array($result)) {

    $imgName = $row["filename"]; 
    $imgPath = $imgLocation . $imgName;


    // Make sure the file exists
    if(!file_exists($imgPath) || !is_file($imgPath)) {
        header('HTTP/1.0 404 Not Found');
        die('The file does not exist');
    }

    // Make sure the file is an image
    $imgData = getimagesize($imgPath);
    if(!$imgData) {
        header('HTTP/1.0 403 Forbidden');
        die('The file you requested is not an image.');
    }


    // Set the appropriate content-type
    // and provide the content-length.

    header("Pragma: public");
    header("Expires: 0");
    header("Cache-Control: must-revalidate, post-check=0, pre-check=0");

    header("Content-Type: image/jpg");
    header("Content-length: " . filesize($imgPath));

    // Print the image data
    readfile($imgPath);
    exit();

}
?>

Answer 1

, QueryString, call_images.php, , , . () .

<?php
// Get our database connector
require("includes/copta.php");

$imgLocation = '/ uploadz/';

$fn = mysql_real_escape_string($_GET['imgPath']);

$sql = "select filename from people WHERE filename = '{$fn}'";

$result = mysql_query($sql) or 
    die ("Could not access DB: " . mysql_error());   

if (mysql_num_rows($result) == 0) {
    header('HTTP/1.0 404 Not Found');
    die('The file does not exist');
}
$imgName = mysql_result($result, 0, 0); 
$imgPath = $imgLocation . $imgName;

// Make sure the file exists
if(!file_exists($imgPath) || !is_file($imgPath)) {
    header('HTTP/1.0 404 Not Found');
    die('The file does not exist');
}

// Make sure the file is an image
$imgData = getimagesize($imgPath);
if(!$imgData) {
    header('HTTP/1.0 403 Forbidden');
    die('The file you requested is not an image.');
}


// Set the appropriate content-type
// and provide the content-length.

header("Pragma: public");
header("Expires: 0");
header("Cache-Control: must-revalidate, post-check=0, pre-check=0");

header("Content-Type: image/jpg");
header("Content-length: " . filesize($imgPath));

// Print the image data
readfile($imgPath);
exit();
?>

:

• $fn = mysql_real_escape_string($_GET['imgPath']); , querystring, , . , , , , ( , - , ).
• ,
• mysql_result(), .
• readfile() fpassthru(), fopen, .