First of all we must connect to our database
<?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
$database = “”; – It’s your database
Now let’s connect to the database
@mysql_connect($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
$id = $_GET['id'];
It will get a id from an URL. blabla.com
So.. your id will be 3 and it will show an image which id is 3
if(!isset($id)Â ||Â 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
$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$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
header('Content-type:Â 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
$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(
$id)Â ||Â empty($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:Â image/jpg');
echo $content;
}
?>
It works perfectly
Enjoy… 🙂 Â Â If you have questions then please feel free to ask
Posted in programming, Tips n Trik | |
how is it done, as mine wont do anything, i get a blank page now, with a box in the middle that says ‘image’…
my code:
if(!isset($id) || empty($id))
{
die(“Please select your image!”);
}
else
{
$query = mysql_query(“SELECT * FROM phpbb_attachments WHERE post_msg_id=’$id'”);
$row = mysql_fetch_array($query);
$content = $row[‘image’];
header(‘Content-type: image/jpg’);
echo $content;
}
i already connect further up in my code, and the $id = $_GET[‘id’];
is up there too..
maybe you can use header(’Content-type: image/gif’); or header(’Content-type: image/bmp’); based on the type of the image that you’ve stored on the database