Can't display image from MySQL database in PHP - php

Question

I have some problems here and I hope someone can help me. unfortunately I couldn't find a solution yet so you are my last hope.
I upload an image to my database but I cant display it. What I get is this icon when an image cant be find or load, but my database is full of pictures :-/
This is my PHP code:
<?php
$dbhost = "localhost";
$dbuser = "DBConnector";
$dbpassword = "root";
$db = "phplogin";
$b = $_POST['bildbestaetigen'];
if(isset($b)){
if(getimagesize($_FILES['bild']['tmp_name'])== FALSE)
{
echo "Choose a picture.";
}
else
{
$image= addslashes($_FILES['bild']['tmp_name']);
$name = addslashes($_FILES['bild']['name']);
$image= file_get_contents($image);
$image= base64_encode($image);
saveimage($name,$image);
}
}
displayimage();
function saveimage ($name,$image){
$link = mysqli_connect("localhost","root","") or die("Verbindung zur Datenbank konnte nicht hergestellt werden!");
mysqli_select_db($link,"phplogin") or die ("kann nich finden ");
$result= mysqli_query($link, "insert into bilder (name,bild) values ('".$name."','".$image."')");
if($result)
{
echo "Image uploaded";
}
else
{
echo "<br/> Image not uploaded";
}
}
function displayimage(){
$link = mysqli_connect("localhost","root","") or die("Verbindung zur Datenbank konnte nicht hergestellt werdengjghghghh");
mysqli_select_db($link,"phplogin") or die ("kann nich finden ");
$result = mysqli_query($link,"SELECT * from bilder");
while( $row = mysqli_fetch_array($result));
{
echo '<img height="300" width="300" src="data:bild;base64,'.$row[2].' ">';
}
mysqli_close($link);
}
?>
I tried so many things with out luck.
I hope someone can help me!

Answer 1

Use this W3Schools tutorial on PHP uploads. As mentioned in comments, save the image path in database, not file itself which may require the blob data type and can bloat your database very quickly.
To do so, upload the user-selected file from your HTML form onto your webserver as a physical file by specifying a target folder with $_FILES array (be sure folder exists) and using the move_uploaded_file() function:
$targetfile = "uploads/" .basename($_FILES['bild'])
...
# FILE VALIDATION
...
move_uploaded_file($_FILES['bild']['tmp_name'], $targetfile)
Then, save this image path (text string) into your database (assuming here bild is a text data type):
insert into bilder (name,bild) values ('".$name."','".$targetfile."')
Finally, just as you did before fetch records from the database table and echo the image path in src attribute of html's img tag (by the way filter the SQL query to particular user or image id as multiple records may output):
echo '<img height="300" width="300" src="'.$row[2].'">';
And remember the $targetfile variable will contain full, absolute path of image location.

Answer 2

Hi, assuming you have correctly uploaded an image into your database
and a folder somewhere inside your server. You can use this code to
print it out.
<?php
//codes to connect to database and mysql query to get image name from table inside database...
$target_dir = "uploads/"; //path of where the images are store on your server
$image_name = $row_image['image_name']; //name of the image retrieved by mysql query
//print the image
echo "<img src=" . "'" . $target_dir . "/" . $image_name . "'" . "style='width:50px;height:50px'/>";
?>