Jason Wong wrote:
On Tuesday 02 November 2004 03:36, P. George wrote:
i know that postgres has a raw binary datatype called bytea for storing
pictures and whatnot, but i'm having trouble finding an example of
displaying an image on a web page with php that is pulled directly from
the database?
On Mon, 2004-11-01 at 21:36 -0600, P. George wrote:
i know that postgres has a raw binary datatype called bytea for storing
pictures and whatnot, but i'm having trouble finding an example of
displaying an image on a web page with php that is pulled directly from
the database?
is this
i know that postgres has a raw binary datatype called bytea for storing
pictures and whatnot, but i'm having trouble finding an example of
displaying an image on a web page with php that is pulled directly from
the database?
is this possible?
if so, how?
thanks.
- philip
--
PHP Database
usually encouraged not to store BLOBs whenever possible due
to the performance hit that's usually associated with that.
-M
-Original Message-
From: P. George [mailto:[EMAIL PROTECTED]
Sent: Monday, November 01, 2004 10:36 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] can i display an image
Typically, it's infinitely more performance-friendly to simply store
your
files on the hard disk and use the database to refer to that location.
For instance, assume your root web dir is /www. Inside there you have
an
/images directory.
If you religiously put all files inside of /images you
On Tuesday 02 November 2004 03:36, P. George wrote:
i know that postgres has a raw binary datatype called bytea for storing
pictures and whatnot, but i'm having trouble finding an example of
displaying an image on a web page with php that is pulled directly from
the database?
is this
Tutorials that shows how to upload and store image/binary files in a
database
also usually shows how to retrieve them (and in the case of images,
also how
to display them).
One function you'll find useful is
imagecreatefromstring()
thanks.
i found an example for uploading an image into the
$result = pg_exec($dbconn, SELECT $image FROM thetable where
name='$name');
nevermind. i had a missing double-quote on this line. works great now.
thanks.
- philip
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