Re: Showing an image
| From: | kachaloo | Date: | Mon, 02 Jul 2001 09:58:02 +0000 |
| Subject: | Re: Showing an image | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-10016@lists.php.net to get a copy of this message | ||
Hi Beau,
The logic makes sense but I don't know how to implement it ?
So pls explain it to me !!
Thanks,
Vishal
My code is :
_______________________________________________________
view.php
<html>
<head>
<title>View the files</title>
</head>
<FORM ENCTYPE="multipart/form-data" ACTION="show.php" METHOD=POST>
<b>ID</b><br><INPUT TYPE="TEXT" NAME="id"
><BR>
<INPUT TYPE="submit" VALUE="Send File">
</form>
</head>
<body>
</body>
</html>
_______________________________________________________
show.php
<?php
$connection=mysql_connect('localhost','root')
or die("Sorry could not connect to the database");
// select database
$db=mysql_select_db('punebuzz',$connection)
or die("Sorry could not connect to the database");
// now the query
$sql="Select * from article where ID='$id'";
$result=mysql_query($sql);
/*
$result = @mysql_query($sql);
$data = @mysql_result($result, 0, "PICTURE");
$type = @mysql_result($result, 0, "FILETYPE");
Header("Content-type: $type");
echo $data;
$cateory = @mysql_result($result, 0, "CATEORY");
$heading = @mysql_result($result, 0, "HEADING");
$body = @mysql_result($result, 0, "BODY");
$keyword = @mysql_result($result, 0, "KEYWORD");
$name = @mysql_result($result, 0, "FILENAME");
$size = @mysql_result($result, 0, "FILESIZE");
$id = @mysql_result($result, 0, "ID");
Header("Content-type: text/html");
echo "<p><b>ID</b>$id</P>";
echo "<p><b>CATEORY</b>$cateory</P>";
echo "<p><b>HEADING</b>$heading</P>";
echo "<p><b>BODY</b>$body</P>";
echo "<p><b>KEYWORD</b>$keyword</P>";
echo "<p><b>FILENAME</b>$filename</P>";
echo "<p><b>FILESIZE</b>$filesize</P>";
*/
while ($row = mysql_fetch_object ($result))
{
$cateory =$row->CATEORY;
$heading =$row->HEADING;
$body =$row->BODY;
$keyword =$row->KEYWORD;
$name =$row->FILENAME;
$size =$row->FILESIZE;
$id =$row->ID;
$picture =$row->PICTURE;
$filetype =$row->FILETYPE;
}
mysql_free_result($result);
mysql_close($db);
?>
_______________________________________________________
----- Original Message -----
From: "Beau Lebens" <B.Lebens@curtin.edu.au>
To: "'kachaloo'" <kachaloo@hotmail.com>; <php-db@lists.php.net>
Sent: Monday, July 02, 2001 1:14 PM
Subject: RE: [PHP-DB] Showing an image
> what you should actually do is save that script as something like
> "image.php" and then on the page where you want to have text + images, do
> something like
>
>
> <html>
> <body>
> <img src="image.php?uid=xxx" border="0"><br>
> I am some text under the image that is grabbed from the DB
> </body>
> </html>
>
> assuming that the image.php page accesses teh DB based on a car called
"uid'
> which represents a key to pull the right image, in this case it is "xxx"
:)
>
> HTH
> Beau
>
>
> // -----Original Message-----
> // From: kachaloo [mailto:kachaloo@hotmail.com]
> // Sent: Monday, 2 July 2001 3:38 PM
> // To: php-db@lists.php.net
> // Subject: [PHP-DB] Showing an image
> //
> //
> // HI,
> // I want to show an image from a database but to do that I am using :
> // $result = @mysql_query($sql);
> // $data = @mysql_result($result, 0, "PICTURE");
> // $type = @mysql_result($result, 0, "FILETYPE");
> // Header("Content-type: $type");
> // echo $data;
> //
> //
> // but after this I also want to show some text info. How can I
> // do this ? I
> // will have to use something else besides header. Please Help.
> // Thanks,
> // Vishal
> //
> //
> //
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