Display all images from database (php) - php

Question

I use this code to upload images in the database:
<?php
//Store the upload form
$UploadForm = " <form id='idForm' action='upload.php' method='post' enctype='multipart/form-data'>
<input type='file' name='image'/><br/><br/>
<input id='BTN' type='submit' value='Upload'/><br/><br/>
</form>";
//if logged in show the upload form
if($userid && $username){
echo $UploadForm;
// Connect to database
$con = mysqli_connect('***', '***', '***', '***_dbimage');
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//file properties
if(isset($_FILES['image'])){
$file = $_FILES['image']['tmp_name'];
}
//if image selected
if(isset($file) && $file != ""){
$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name = addslashes($_FILES['image']['name']);
$image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size == FALSE){
echo "That's not an image!";
header( "refresh:2;url=upload.php" );
}
else{
$qry = mysqli_query($con,"SELECT * FROM store WHERE name='$image_name'");
$Nrows = $qry->num_rows;
if( $Nrows == 0){
if(!$insert = mysqli_query($con,"INSERT INTO store VALUES ('','$image_name','$username','$image')")){
echo "We had problems uploading your file!";
header( "refresh:2;url=upload.php" );
}
else{
echo "Image $image_name uploaded!";
header( "refresh:2;url=upload.php" );
}
}
else{
echo "There is already an image uploaded with the name $image_name<br/>";
}
}
}
else{
echo "Please select an image";
}
mysqli_close($con);
}
else{
echo "You have to be logged in to upload!";
}
?>
And this code to display all images from the database:
// Connect to database
$con = mysqli_connect('***', '***', '***', '***_dbimage');
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$res = mysqli_query($con,'SELECT * FROM store');
while($row = $res->fetch_array()){
$image = $rows['image'];
echo "<img src='".$image."' />";
}
And I get something like tons of random symbols like diamonds with question marks in them and letters instead of my image. The scripts are not made by me. I just watched some tutorials and combined them and it seems that I didn't "combined" them properly. What am I doing wrong?
LATER EDIT:
HTML:
<img src="getImage.php?id=26"/>
PHP (getImage.php):
$con = mysqli_connect('***', '***', '***', '***_dbimage');
if(isset($_GET['id']))
{
$id = mysql_real_escape_string($_GET['id']);
$query = mysql_query("SELECT * FROM store WHERE id=$id");
while($row = mysql_fetch_assoc($query))
{
$imageData = $row['image'];
}
header("content-type:image/jpeg");
echo $imageData;
}
else
{
echo "Error!";
}
?>
Still can't get it to work! Help please!

Answer 1

I finaly did it!
This is the upload script:
<?php
//Store the upload form
$UploadForm = " <form id='idForm' action='upload.php' method='post' enctype='multipart/form-data'>
<input type='file' name='image'/><br/><br/>
<input id='BTN' type='submit' value='Upload'/><br/><br/>
</form>";
//if logged in show the upload form
if($userid && $username){
echo $UploadForm;
// Connect to database
$con = mysqli_connect('***', '***', '***', '***_dbimage');
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//file properties
if(isset($_FILES['image'])){
$file = $_FILES['image']['tmp_name'];
}
//if image selected
if(isset($file) && $file != ""){
$image = mysqli_real_escape_string($con,file_get_contents($_FILES['image']['tmp_name']));
$image_name = addslashes($_FILES['image']['name']);
$image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size == FALSE){
echo "That's not an image!";
header( "refresh:2;url=upload.php" );
}
else{
$qry = mysqli_query($con,"SELECT * FROM store WHERE name='$image_name'");
$Nrows = $qry->num_rows;
if( $Nrows == 0){
if(!$insert = mysqli_query($con,"INSERT INTO store VALUES ('','$image_name','$username','$image')")){
echo "We had problems uploading your file!";
header( "refresh:2;url=upload.php" );
}
else{
echo "Image $image_name uploaded!";
header( "refresh:2;url=upload.php" );
}
}
else{
echo "There is already an image uploaded with the name $image_name<br/>";
}
}
}
else{
echo "Please select an image";
}
mysqli_close($con);
}
else{
echo "You have to be logged in to upload!";
}
?>
Here is the diplay script:
$con = mysqli_connect('***', '***', '***', '***_dbimage');
$query = mysqli_query($con,"SELECT id FROM store");
while($row = mysqli_fetch_assoc($query))
{
$IDstore = $row['id'];
echo "<img src='getImage.php?id=".$IDstore."'/>";
}
And the "getImage.php":
<?php
$con = mysqli_connect('***', '***', '***', '***_dbimage');
if(isset($_GET['id']))
{
$id = mysqli_real_escape_string($con,$_GET['id']);
$query = mysqli_query($con,"SELECT * FROM store WHERE id=$id");
while($row = mysqli_fetch_assoc($query))
{
$imageData = $row['image'];
}
header("content-type:image/jpeg");
echo $imageData;
}
else
{
echo "Error!";
}
?>
I hope it will help someone cause it's ready to use now. :)