Okay..I'm going to show y'all how to show your images.. those which are in your DB
It's very very simple.. so..
First of all we must connect to our database
PHP Code:
<?php
$username = "";
$password = "";
$host = "localhost";
$database = "";
$username = ""; - It's your database's username
$password = ""; - It's your database's password
$host = "localhost"; - It's your database's host, usually it's localhost but it can be something else also
$database = ""; - It's your database
Now let's connect to the database
PHP Code:
@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());
@mysql_select_db($database) or die("Can not select the database: ".mysql_error());
@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error()); - This connects to your database
@mysql_select_db($database) or die("Can not select the database: ".mysql_error()); - This will select your database
Now let's get our id
PHP Code:
$id = $_GET['id'];
It will get a id from an URL.
blabla.com
So.. your id will be 3 and it will show an image which id is 3
PHP Code:
if(!isset($id) || empty($id)){
die("Please select your image!");
}else{
if(!isset($id) || empty($id)){ - If this ID in your url is empty like
?id= of if it's not even set
die("Please select your image!"); - lets display an error
}else{ - But if it is set let's continue with showing our image
PHP Code:
$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'"); - Now let's select the blob from the table where id is $id (for example: 3)
$row = mysql_fetch_array($query); - Let's gather our info about image which id is $id into one variable
$content = $row['image']; - Get's the blob from our table
Now let's display our image
PHP Code:
header('Content-type: image/jpg');
echo $content;
}
header('Content-type: image/jpg'); - This tells to the browser and to the server that this file will be a jpg file
echo $content; - This will display our blob..
} - Ends else
Okay.. now here's our full script.
PHP Code:
<?php
$username = "";
$password = "";
$host = "localhost";
$database = "";
@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());
@mysql_select_db($database) or die("Can not select the database: ".mysql_error());
$id = $_GET['id'];
if(!isset($id) || empty($id)){
die("Please select your image!");
}else{
$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header('Content-type: image/jpg');
echo $content;
}
?>
It works perfectly
Enjoy
If you have questions then please feel free to ask