php script does not show all images stored in a table - php

Question

in mysql Datbase there is images stored using php Script (image got from a form.html/POST method) let's cal them (phpImages). and there is others stored using android application ( by converting Bitmap to String and using StringBuilder ). let's call them (androidImages).
with this php script i can load and display phpImages, but i cannot display androidImages.
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT image FROM images WHERE id = '$id'";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
echo ( $result['image']);
mysqli_close($con);
}
?>
with this php script i can load androidImages, but i cannot load phpImages :
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT image FROM images WHERE id = '$id'";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
echo base64_decode( $result['image'] );
mysqli_close($con);
}
?>
i wan't a php script that could display the both. because i want to load all images in a ListView of an android Apps.
**This is php script relied to android Application : **
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
$image = $_POST['image'];
$con=mysqli_connect("localhost","root","","othmane")or die(mysqli_error($con));
$sql = "INSERT INTO images (image,image_type) VALUES (?,'android')";
$stmt = mysqli_prepare($con,$sql);
mysqli_stmt_bind_param($stmt,"s",$image);
mysqli_stmt_execute($stmt);
$check = mysqli_stmt_affected_rows($stmt);
if($check == 1){
echo "Image Uploaded Successfully";
}else{
echo "Error Uploading Image";
}
mysqli_close($con);
}else{
echo "Error";
}
?>
this is php Script relied with Form.html post method :
<?php
echo ini_get( 'file_uploads' );
if(!isset($_POST['submit'])){
echo '<p>Please Select Image to Upload</p>';
}
else
{
try {
upload();
}
catch(Exception $e)
{
echo '<h4>'.$e->getMessage().'</h4>';
}
}
function upload(){
$imgfp = fopen($_FILES['photo']['tmp_name'], 'rb');
print_r($_FILES);
$dbh = new PDO("mysql:host=localhost;dbname=othmane", 'root', '');
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $dbh->prepare("INSERT INTO images (image,image_type) VALUES (?,'php')");
$stmt->bindParam(1, $imgfp, PDO::PARAM_LOB);
$stmt->execute();
}
?>

Answer 1

Add a column called image_type in your table and pass one of the following values to determine what the source of the image is upon uploading: phpImage or androidImage
So you can do:
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if ($_SERVER['REQUEST_METHOD'] == 'GET'){
$id = $_GET['id'];
$sql = "SELECT image, image_type FROM images WHERE id = '$id'";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));
$result = mysqli_fetch_array($r);
header('Content-Type: image/jpeg');
if ($result['image_type'] == 'phpImage') {
echo ( $result['image']);
} else if ($result['image_type'] == 'androidImage') {
echo base64_decode( $result['image'] );
}
mysqli_close($con);
}
?>