Answer Posted / vinod patil

Just to polish the topic off, below are the two files that
have been tested
on my machine (xp, apache2, php 4.3.9, mysql4.0.1x)

<?php
//show_image.php

require("conn.php");
//check to see if the id is passed
if(isset($_GET['id'])) {
$id=$_GET['id'];


$query = "select bin_data, filetype from binary_data where
id=$id";
//echo $query;
$result = connect($query);
$row = mysql_fetch_array($result);
{
$data = $row['bin_data'];
$type = $row['filetype'];
}
if ($type=="pjpeg") $type = "jpeg";
Header( "Content-type: $type");

echo $data;

}
?>

<?php

// show_desc.php
require("conn.php");

// you may have to modify login information for your
database server

$query = "select description, id from binary_data ";
$result = connect($query);

while ($rows = MYSQL_FETCH_ARRAY($result))
{
echo $rows['description'];
echo "<br><br>";

echo "<img src=\"show_image.php?id=".$rows['id']."\">\n";


};
?>

And I used table structure

CREATE TABLE binary_data (
id INT(4) NOT NULL AUTO_INCREMENT PRIMARY KEY,
description CHAR(50),
bin_data LONGBLOB,
filename CHAR(50),
filesize CHAR(50),
filetype CHAR(50)
);

hth

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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In .net how many error will occur?

2422

Program ended with error or warning, return code: 4

13698