Posted by ste on 04/28/06 15:55

"Jerry Stuckle" <jstucklex@attglobal.net> wrote in message
news:G5udneifg5pfZ8zZnZ2dnUVZ_tGdnZ2d@comcast.com...
<snip>
> You're close. mysql_query() returns a result object; you need to retrieve
> the actual data (into $row in your example).
>
> After the mysql_query() add the following:
>
> $row = mysql_fetch_array($result);
>
>
> --
> ==================
> Remove the "x" from my email address
> Jerry Stuckle
> JDS Computer Training Corp.
> jstucklex@attglobal.net
> ==================

Hi Jerry,

Thanks for getting back to me - I'm close? I'm astonished as I thought I
would be way out!

I *think* I've put your line of code in the right place, but this still
doesn't work I'm afraid. When the page opens, there's no image - if I look
at the HTML, it is basically returning <img src=""> instead of <img
src="images/image1/jpg">. Perhaps the query is right but my code to insert
the field name (which contains the image URL, it's called 'imagelocation')
isn't?

Here's the code again with the extra line - did I put it in the right place?

<?php
include("my_db_login.inc");

$connection = mysql_connect($host,$user,$password) or die ("couldn't connect
to server");
$db = mysql_select_db($database,$connection) or die ("Couldn't select
database");

$query = "SELECT * FROM my_database WHERE imageid =
\"{$_POST['imageid']}\"";
$result = mysql_query($query) or die ("Couldn't execute query.");
$row = mysql_fetch_array($result);

echo "<img src=\"{$row['imagelocation']}\" width=\"500\" border=\"0\" />";
?>

Thanks,

Ste

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