On Tuesday 13 March 2007 22:09, Tijnema ! wrote:
> 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!.
He's not using image.php to insert. Earlier he mentioned using phpmyadmin to
insert the image, that was the way I used too. First learn to display an
image, this way its easier to know if any upload script you make up later is
working correctly.
>
> 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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Kennel Arivene
http://www.arivene.net
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