Mika,
Put the dollar sign (i.e., $) outside the curly brace.
$query="SELECT * FROM pic_upload WHERE band_id='${band_id}'";
A-
-----Original Message-----
From: Mika Jaaksi [mailto:mika.jaa...@gmail.com]
Sent: Thursday, February 12, 2009 12:27 PM
To: php-db@lists.php.net
Subject: [PHP-DB] Re: session variable in select query showing picture
from database
Still fighting with it...
So, these work:
$query="SELECT * FROM pic_upload;
$query="SELECT * FROM pic_upload WHERE band_id=11";
picture is shown on the other page
but when adding variable into query it doesn't show the picture on the
other
page
$query="SELECT * FROM pic_upload WHERE band_id='{$band_id}'";
I'm out of ideas at the moment...
ps. forget what I said about the weird markings...
2009/2/12 Mika Jaaksi <mika.jaa...@gmail.com>
> 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...
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