Answer / vikaskaundal40@gmail.com

firstly img convert to binery and which uesd in file stream
and read the file access mode and again you covert the img
in bytes and again used in file stream and write and
retrive the img and you upload all type img in very easly

Answer / 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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To unsubscribe, visit: http://www.php.net/unsub.php

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