display list of images on the browser using php - php

Question

I'm trying to display on the browser 3 images stored in mysql database , but i figured out that just the first image of the first iteration on loop WHILE is displayed. how to display the 3 images in the browser ?
<?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 between 2 and 6";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
$ss=$result['image_type'];
while($row = $r->fetch_assoc()){
if ($ss == 'php') {
echo ( $result['image']);
} else if ($ss == 'android') {
echo base64_decode( $result['image'] );
}
}
}else{
mysqli_close($con);
}
?>
this code :
<?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 between 1 and 3";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
//$result=mysqli_fetch_array($r);
$result=var_dump(mysqli_fetch_all($r));
//header('Content-Type:image/jpeg');
while($row = mysqli_fetch_assoc($r))
{
var_dump($row) ;
}
/*while($row = $r->fetch_assoc()){
$ss=$row['image_type'];
if ($ss == 'php') {
echo ( $row['image']);
} else if ($ss == 'android') {
echo base64_decode( $row['image'] );
}
}*/
}
else{
mysqli_close($con);
}
?>
and this is the result :

Answer 1

edit. as i've looked throught the problem, you should be able to do:
<img src="data:image/jpeg;base64,
<?php echo base64_encode( $image_data ); ?>
" />
to display all of your image, so below should work:
while($row = mysqli_fetch_assoc($r)){
$ss = $row['image_type'];
if ($ss == 'php') {
<img src="data:image/jpeg;base64, <?php echo ( $row['image']); ?>" />
} else if ($ss == 'android') {
<img src="data:image/jpeg;base64, <?php echo base64_decode( $row['image'] ); ?> " />
}
}
When you are doing $r->fetch_assoc() ( so as long as your query is getting you result, your While loop will go on), you are fetching one result row of your data into associative array, so you have your data in $row variable.
You have to use that variable for checking your ['image_type'] and getting your ['image'].
In your code your $result['image_type'] and $result['image'] was bound to the first record that you have fetched with $result=mysqli_fetch_array($r) ( as documentation states: mysqli_result::fetch_array -- mysqli_fetch_array — Fetch a result row as an associative, a numeric array, or both ).
if($_SERVER['REQUEST_METHOD']=='GET'){
$sql = "SELECT image,image_type FROM images where id BETWEEN 2 and 6";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));
header("Content-type: image/jpeg");
while($row = mysqli_fetch_assoc($r)){
$ss = $row['image_type'];
if ($ss == 'php') {
echo ( $row['image']);
} else if ($ss == 'android') {
echo base64_decode( $row['image'] );
}
}
}else{
mysqli_close($con);
}
also when you are using procedular style, use it all the way, as (examples from php.net):
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_assoc($result)) {
printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}
or just use object oriented style:
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";
$result = $mysqli->query($query);
while ($row = $result->fetch_assoc()) {
printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}
More info, examples here: mysqli_result::fetch_assoc