How can i display image from database? - php

Question

Image can't display from database in php and it's fire such error.
When I execute this code such error is occure The image cannot be display beacuse it's contain error. Image is store in mysql database and field type is BLOB. IT didn't find any solution to avoid this error.
<?php
require_once('database/db.class.php');
$objDatabase=new db_class();
$objDatabase->connect();
$sql="SELECT id,firstname,lastname,sex,email,userimage,ext,city,state,username FROM t_users WHERE username='".$_SESSION['username']."'";
$totalRow=$objDatabase->select($sql);
if($row=$objDatabase->get_row($totalRow, 'MYSQL_BOTH'))
{
$userArray=array('id'=>$row['id'],'firstname'=>$row['firstname'],'lastname'=>$row['lastname'],'sex'=>$row['sex'],'email'=>$row['email'],'userimage'=>$row['userimage'],'ext'=>$row['ext'],'city'=>$row['city'],'state'=>$row['state'],'username'=>$row['username']);
}
?>
<table cellspacing="10px" width="100%">
<tr>
<td width="23%" style="vertical-align:top">
<?php
header("Content-type: image/{$userArray['ext']}");
header("Content-Length: " . strlen($userArray['userimage']));
echo $userArray['userimage'];
?>
</td>
</tr>
</table>
Upload Image Code
#list(, , $imtype, ) = getimagesize($_FILES['photo']['tmp_name']);
// Get image type.
// We use # to omit errors
if ($imtype == 3) // cheking image type
$ext="png"; // to use it later in HTTP headers
elseif ($imtype == 2)
$ext="jpeg";
elseif ($imtype == 1)
$ext="gif";
else
$msg = 'Error: unknown file format';
$img = file_get_contents($_FILES['photo']['tmp_name']);
$img = mysql_real_escape_string($img);
// Preparing data to be used in MySQL query
$data=array('firstname'=>$_REQUEST['firstname'],'lastname'=>$_REQUEST['lastname'],'sex'=>$_REQUEST['gender'],'email'=>$_REQUEST['email'],'userimage'=>$img,'ext'=>$ext,'city'=>$_REQUEST['city'],'state'=>$_REQUEST['state'],'zipcode'=>$_REQUEST['zipcode'],'username'=>$_REQUEST['usernameText'],'password'=>$_REQUEST['passwordText']);
$result=$objDatabase->insert_array('t_users',$data);
<input name="photo" type="file" id="photo">

Answer 1

edit - I forgot to include: base64 files will be considerably larger than BLOB.
Try saving it as base64 instead of BLOB.
Here is a function I found on http://www.php.net/manual/en/function.base64-encode.php
function base64_encode_image ($filename=string,$filetype=string) {
if ($filename) {
$imgbinary = fread(fopen($filename, "r"), filesize($filename));
return 'data:image/' . $filetype . ';base64,' . base64_encode($imgbinary);
}
}
When you retrieve the image from the database, it will be as a base64 string, you can put the string directly into the src="" attribute in your HTML (no need to decode). Browsers can parse base64 and display an image.

Answer 2

Well, this isn't the way to do it, just store the image as per usual, then store the path in the database.
If you do want to do it this way, you'll need to make a new file, called something like image.php, then put
<?php
require_once('database/db.class.php');
$objDatabase=new db_class();
$objDatabase->connect();
$sql="SELECT id,firstname,lastname,sex,email,userimage,ext,city,state,username FROM t_users WHERE username='".$_SESSION['username']."'";
$totalRow=$objDatabase->select($sql);
if($row=$objDatabase->get_row($totalRow, 'MYSQL_BOTH'))
{
$userArray=array('id'=>$row['id'],'firstname'=>$row['firstname'],'lastname'=>$row['lastname'],'sex'=>$row['sex'],'email'=>$row['email'],'userimage'=>$row['userimage'],'ext'=>$row['ext'],'city'=>$row['city'],'state'=>$row['state'],'username'=>$row['username']);
}
header("Content-type: image/{$userArray['ext']}");
header("Content-Length: " . strlen($userArray['userimage']));
echo $userArray['userimage'];
?>
in to that file.
Then in the table use
<img src='/image.php ?>' alt='' />
But don't, just do it by storing the path in the DB, it's rare that you should be storing binary data in a database that's not the file system, for reasons I won't go into here.