RE: [PHP-DB] storing images in database

And how are you feeding the $id?....are you setting a value for that element? In the sample code the default is the record_id that corresponds back to the id of the row with the image blob field. Bastien

From: Chip Wiegand <chip.wiegand@simrad.com> To: bastien_k@hotmail.com CC: "PHP DB" <php-db@lists.php.net> Subject: RE: [PHP-DB] storing images in database Date: Tue, 25 Jan 2005 12:57:40 -0800 "Bastien Koert" <bastien_k@hotmail.com> wrote on 01/25/2005 12:46:12 PM: yes goes back to the whole header problem which is why you are here. If you could post the code, it would be simpler to help you... Bastien This is in the main page - <? printf("<p><img src=\"image-src.php?id=$id\" alt=\"hotspot images\">%s</p>", $row["text"]); ?> and this is in a new included page - <? if($_GET['id']) { $id = $_GET['id']; $query = "select * from hotspots where id=$id"; $result = @MYSQL_QUERY($query); $data = @MYSQL_RESULT($result,0,"image"); $type = @MYSQL_RESULT($result,0,"type"); Header( "Content-type: $type"); echo $data; }; ?> The database connection statements are in an include file called at the top of the main page. In the first statement shown above the alt text for the image appears on the web page just fine, the image itself shows a broken image icon. FWIW, I have the image stored in the database in a blob field, is that correct? -- Chip

From: Chip Wiegand <chip.wiegand@simrad.com> To: bastien_k@hotmail.com Subject: RE: [PHP-DB] storing images in database Date: Tue, 25 Jan 2005 12:44:44 -0800 "Bastien Koert" <bastien_k@hotmail.com> wrote on 01/25/2005 12:39:15

PM:

Its not src='id=$id'> that will defnintely blow up.... echo '<img src="./path/to/image.php?id=$id">'; where $id is the id of the record you are trying to get the image

to...

Bastien

So the code has to be a seperate included page I guess? -- Chip

From: Chip Wiegand <chip.wiegand@simrad.com> To: bastien_k@hotmail.com CC: php-db@lists.php.net Subject: RE: [PHP-DB] storing images in database Date: Tue, 25 Jan 2005 12:37:15 -0800 Thanks Bastien, In testing this I have added the code samples to a page and have it working except the path statement is not correct. For now, I've

just

added

all the code to one page, rather than including a second page. The statement - echo '<img src="id=$id">'; is resulting in this error -

The

requested URL /id=$id was not found on this server. Any

suggestions?

Thanks, Chip "Bastien Koert" <bastien_k@hotmail.com> wrote on 01/25/2005

09:45:39

AM:

the best way to do this is to move the image processing code to

a

separate

page and include it like this echo '<img src="./path/to/image.php?id=$id">'; then the image page looks like this: <?php if($_GET['id']) { $id = $_GET['id']; // you may have to modify login information for your database

server:

@MYSQL_CONNECT("localhost","root","password"); @mysql_select_db("binary_data"); $query = "select bin_data,filetype from binary_data where

id=$id";

$result = @MYSQL_QUERY($query); $data = @MYSQL_RESULT($result,0,"bin_data"); $type = @MYSQL_RESULT($result,0,"filetype"); Header( "Content-type: $type"); echo $data; }; ?> bastien

From: Chip Wiegand <chip.wiegand@simrad.com> To: "PHP DB" <php-db@lists.php.net> Subject: [PHP-DB] storing images in database Date: Tue, 25 Jan 2005 09:11:07 -0800 I have stored a .jpg image in a database, then when I make a

sql

statement

to display that image on a web page all I get is the cryptic

code

in

place

of the image. I am storing it in a row configured as a blob,

mime

type

image/jpeg and binary (using phpMyAdmin). What am I doing

wrong?

Regards, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

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