storing images in a database by karan chanana
TRANSCRIPT
Storing images in a database
<?PHP
$dbcnx = @MYSQL_CONNECT("localhost", "username",
"password");
IF (!$dbcnx) {
ECHO( "<p>connection to database server failed!</p>"
);
EXIT();
}
IF (! @MYSQL_SELECT_DB("base64imgdb") ) {
ECHO( "<p>Image Database Not Available!</p >" );
EXIT();
}
?>
The Image Reader IMAGE.PHP
This file may be the hardest file to understand whenever you see
how simple view.php is, but bear with me, your patience will pay
off. This file takes a request, requests the row in the table,
decodes the data, and presents itself as an image. First, we
have to connect to the database again:
<?PHP
$dbcnx = @MYSQL_CONNECT("localhost", "username",
"password");
IF (!$dbcnx) {
ECHO( "
connection to database server failed!
"
);
EXIT();
}
IF (! @MYSQL_SELECT_DB("base64imgdb") ) {
ECHO( "
Image Database Not Available!
" );
EXIT();
}
?>
Now we need to find out which row it's requesting, which is done
using image.php?img=x:
<?PHP
$img = $_REQUEST["img"];
?>
After this, we need to connect to the table, get the data, and
set it into variables:
<?PHP
$result = @MYSQL_QUERY("SELECT * FROM images WHERE imgid=" . $img .
"");
IF (!$result) {
ECHO("
Error performing query: " . MYSQL_ERROR()
. "
");
EXIT();
}
WHILE ( $row = MYSQL_FETCH_ARRAY($result) ) {
$imgid = $row["imgid"];
$encodeddata = $row["sixfourdata"];
}
?>
Now here is the last and most confusing part of the file:
<?PHP
ECHO BASE64_DECODE($encodeddata);
?>
Now let me explain this. All this does is decodes the base64-encoded
image data, end echos it. That's it, nothing else.
VIEW.PHP (example viewer)
Okay, so you made it this far already. This is now the easiest
to copy and paste but hardest part to understand, where
image.php?img=1 matches with whatever row the image is on, for
example if it's row 357 then you would need to put
image.php?img=357:
<img src='image.php?img=1' border="0" alt="">
Now that wasn't so hard was it? But most of you are probably
wondering why when you link to a page, you get an image. This
is the reason: images arent defined by their 3 letter suffixes
(such as jpg or gif), but by how their headers are written.
IMAGE.PHP simply echos the image data, and acts like an image
even though it just proccesses the request. This is why you get
an image.
readdir.php:
<?PHP
###############################
# DB CONNECTION
# CHANGE THESE VALUES
###############################
$dbcnx = @MYSQL_CONNECT("localhost", "username",
"password");
IF (!$dbcnx) {
ECHO( "
connection to database server failed!
"
);
EXIT();
}
IF (! @MYSQL_SELECT_DB("base64imgdb") ) {
ECHO( "
Image Database Not Available!
" );
EXIT();
}
$path = "./";
$dir_handle = @OPENDIR($path) or DIE("Unable to open directory $path");
WHILE ($file = READDIR($dir_handle)) {
$filetyp = SUBSTR($file, -3);
IF ($filetyp == 'gif' OR $filetyp == 'jpg') {
$handle = FOPEN($file,'r');
$file_content = FREAD($handle,FILESIZE($file));
FCLOSE($handle);
$encoded = CHUNK_SPLIT(BASE64_ENCODE($file_content));
$sql = "INSERT INTO images SET sixfourdata='$encoded'";
@MYSQL_QUERY($sql);
}
}
CLOSEDIR($dir_handle);
ECHO("complete");
?>
image.php:
<?PHP
$dbcnx = @MYSQL_CONNECT("localhost", "username",
"password");
IF (!$dbcnx) {
ECHO( "
connection to database server failed!
"
);
EXIT();
}
IF (! @MYSQL_SELECT_DB("base64imgdb") ) {
ECHO( "
Image Database Not Available!
" );
EXIT();
}
$img = $_REQUEST["img"];
$result = @MYSQL_QUERY("SELECT * FROM images WHERE imgid=" . $img .
"");
IF (!$result) {
ECHO("
Error performing query: " . MYSQL_ERROR()
. "
");
EXIT();
}
WHILE ( $row = MYSQL_FETCH_ARRAY($result) ) {
$imgid = $row["imgid"];
$encodeddata = $row["sixfourdata"];
}
ECHO BASE64_DECODE($encodeddata);
?>
view.php:
<html>
<body>
..
<img src='image.php?img=1' border="0" alt="">
..
</body>
</html>