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.
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