On 3/13/07, Bruce Gilbert <webguync@xxxxxxxxx> wrote:
On 3/13/07, Tijnema ! <tijnema@xxxxxxxxx> wrote:
> So you just need to set the content-type and output
> add this to the bottom of the script:
> header("Content-Type: ".$encodeddata);
> echo $title;
>
> If i understand you right.
>
> Tijnema
>
> >
Thanks,
I changed the code around some and now have:
[php]
<?php
//check for validity of user
$db_name="bruceg_mailinglist";
$table_name ="image_holder";
$connection = @mysql_connect("db_address", "uasername", "password")
or die (mysql_error());
$db = @mysql_select_db($db_name, $connection) or die (mysql_error());
$img = $_REQUEST["img"];
$result = @mysql_query("SELECT * FROM image_holder WHERE id=" . $img . "");
if (!$result)
{
echo("Error performing query: " . mysql_error() . "");
exit();
}
while ( $row = @mysql_fetch_array($result) )
{
$imgid = $row["id"];
header("Content-Type: ".$encodeddata);
echo $title;}
?>
[/php]
and in the HTML
<center><img src="image.php?id=1" width="200" border="1" alt=""></center>
but I am getting a MySQL error
"Error performing query: You have an error in your SQL syntax; check
the manual that corresponds to your MySQL server version for the right
syntax to use near '' at line 1"
--
::Bruce::
You changed your html code, you have id=1, and in your PHP code you
are requesting img, so change
<center><img src="image.php?id=1" width="200" border="1" alt=""></center>
to
<center><img src="image.php?img=1" width="200" border="1" alt=""></center>
But i must also say, it is NOT safe to input data from ?img= directly
into your database, someone could do a SQL injection right away with
this code!.
Then about this piece of code
while ( $row = @mysql_fetch_array($result) )
{
$imgid = $row["id"];
header("Content-Type: ".$encodeddata);
echo $title;
}
I hope for you that there's only one item with this id, if not, there
would come an error again, so a while loop is not needed, and second,
now you don't define $encodeddata and $title anymore, try this piece
of code instead of the one above:
$row = @mysql_fetch_array($result);
header("Content-Type: ".row['mimetype']);
echo $row['filecontents'];
ps. Reply to the full PHP list, not just me...
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