Yes SIR! Steve!
THIS IS WORKING NOW!
MANY... MANY... MANY... THANKS FOR YOUR CODE!
As a newbie... I have been working so hard on this one...
You noticed that I have a lot of reading to do...
Best regards,
Merci encore, (Thanks again,)
Yves
At 22:41 -0500 12/02/2001, Steve Werby wrote:
>"Malouin Design Graphique" <[EMAIL PROTECTED]> wrote:
>> Is there anything wrong with this code below?
>> Any help or fix to this code would be very much appreciated.
>
>Let's take a look.
>
>> $db = mysql_connect("www.server.com", "root", "password");
>> mysql_select_db("db_name", $db);
>> $sql = "select * from table_name order by 'date'";
>> $sql = "select * from table_name";
>
>The second $sql overwrites the first so hopefully you didn't want the first.
>
>> $result = mysql_query($sql);
>> while ($row = mysql_fetch_array($result)) {
>> print("<tr><td bgcolor=\"#003399\">");
>> printf("<img src=\"$indice_url\">%s</td></tr>\n",
>> $row["indice_url"]);
>
>Where is $indice_url being set? You call it in printf() but it was not
>previously set in your code. I think you just don't understand how to
>properly use printf() and since you're not really using it to format a
>string, like it's intended to be used I'm not going to bother to explain how
>to use it. I have an idea what you're trying to do, but in my opinion
>printf(%s, $var) is a horrible way to pull data from a query result so
>instead of showing you how to fix it, I'll show you a better way. Why not
>do something like:
>
>$result = mysql_query( $sql );
>while ( $row = mysql_fetch_array( $result ) )
> {
> echo "<tr><td><img src=\"$row[indice_url]\">$row[indice]</td></tr>\n";
> }
>
>--
>Steve Werby
>COO
>24-7 Computer Services, LLC
>Tel: 804.817.2470
>http://www.247computing.com/
--
------------------------
Malouin Design Graphique
http://www.malouin.qc.ca
Québec (Québec) CANADA
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