how can I retrieve the image from folder for a specific person?

Question

I'm trying to save my images to a folder inside the root folder and the name of the image in the database. Now, how am I going to retrieve the specific image for the particular id ?

In my codes, I can only retrieve all the images but I only want one image base on his id and image_name

Is that possible?

Here is my codes:

Here is the codes for uploading the image to the folder and the image_name to the database:

<?php

include '../session.php';
require_once 'config.php';


if (isset($_POST['submit'])) {

    $img_dir = "../updated_photo/";
    $target_file = $img_dir . basename($_FILES["image"]["name"]);
    $imageName = $_FILES["image"]["name"];
    $imageData = file_get_contents($_FILES["image"]["tmp_name"]);
    $imageType = $_FILES["image"]["type"];

        if (substr($imageType, 0,5) == "image") {
            $query = "UPDATE `crew_info` SET `image_name` = ? WHERE `id` = ?";
            $stmt = mysqli_prepare($conn, $query);
            mysqli_stmt_bind_param($stmt, 'si', $imageName, $_POST['id']);
            mysqli_stmt_execute($stmt);
            file_put_contents($target_file, $imageData);
            $id = $_POST['id'];
            header("Location: ../admin/view_all_info.php?id=$id");

        }

        else {

            echo "Image not Uploaded!";

        }

}

?>

Below is the code for retrieving the image and the image_name

<?php 

include '../session.php'; 
require_once 'config.php'; 

if (isset($_REQUEST['id'])) { 

    $query = "SELECT `id`, `image_name` FROM `crew_info` WHERE `id` = ?"; 
    $stmt = mysqli_prepare($conn, $query); 
    mysqli_stmt_bind_param($stmt, 'i', $_REQUEST['id']); 
    mysqli_stmt_execute($stmt); 
    mysqli_stmt_bind_result($stmt, $id, $image_name); 

        while (mysqli_stmt_fetch($stmt)) { 

            echo sprintf("%s", $image_name);
        }

        $img_dir = "../updated_photo/";
        $images = glob($img_dir."*.jpg");

        foreach ($images as $img) {

           echo '<img src="'.$img.'" />';

        } 
} 
else { 
    echo "No Image"; 


}


?>

In my retrieving code, I've tried to use if statement instead of foreach like the code below

if (count($images) > 0) {

   $img = $images[0];
   echo '<img src="'.$img.'" />';

} 

Answer 1

This statement will fetch all the jpg photos from that directory

$img_dir = "../updated_photo/";
        $images = glob($img_dir."*.jpg");

        foreach ($images as $img) {

           echo '<img src="'.$img.'" />';

        } 

If you want to display just one photo you can do this after getting image_name from database

$img_dir = "../updated_photo/";
$img_path = $img_dir.$image_name;
 echo '<img src="'.$img_path.'" />';