Re: [PHP-DB] session variable in select query showing picture from database

On Feb 12, 2009, at 6:07 AM, Mika Jaaksi wrote:

> I'm trying to show picture from database. Everything works until I add
> variable into where part of the query.
>
> It works with plain number. example ...WHERE id=11... ...picture is  
> shown on
> the page.
>
> Here's the code that retrieves the picture. show_pic.php
>
> <?php
> function db_connect($host='********', $user='********',
> $password='********', $db='********')
> {
> mysql_connect($host, $user, $password) or die('I cannot connect to  
> db: ' .
> mysql_error());
> mysql_select_db($db);
> }
> db_connect();
> $band_id = $_SESSION['session_var'];
> $query="SELECT * FROM pic_upload WHERE band_id=$band_id";
> $result=mysql_query($query);
> while($row = mysql_fetch_array($result))
> {
> $bytes = $row['pic_content'];
> }
> header("Content-type: image/jpeg");
> print $bytes;
>
>
> exit ();
> mysql_close();
> ?>
>
>
> other page that shows the picture
>
> <?php
> echo "<img width='400px' src='./show_pic.php' />";
> ?>
>
> Any help would be appreciated...

I'm not positive... But I believe you need to  enclose $band_id like  
this:

$query = "SELECT * FROM pic_upload WHERE band_id='{$band_id}'";

Give that a shot, and if it doesn't work, post back and we'll take it  
from there.