RE: [PHP-DB] storing images in database

Filesystem is easier, by far since it avoids the content header. Glad you solved it 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 14:32:11 -0800 I have done it an easier way, and probably a better way all-around anyway. I am storing the images in a directory and have the script call the file/alt text/title text and a description text in a paragraph below the image. It works quite well this way. What I'm doing is on this home page there is a place for a 'hotspot', a special mention area for the latest news about a particular item. So I have a directory and in it will store an image file called 'latest.gif', so any new image that gets put here will overwrite the existing image. This will be fine for the purposes and the site. Here is the code I have used - <? include "connect"; $sql="select * from hotspots"; $result = mysql_query($sql); while ( $row = mysql_fetch_array($result)) { printf("<p><img src=\"hotspots/latest.gif\" alt=\"%s\" title=\"%s\"></p><p class=\"footnote\">%s</p>", $row["alt-text"], $row["title-text"], $row["desc-text"]); } ?> I'm sure there are many ways to do this sort of thing, but this is quick and easy, and works. Thanks guys, -- Chip "Bastien Koert" <bastien_k@hotmail.com> wrote on 01/25/2005 01:06:01 PM: 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

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

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