[DJDrew22200]

Hi there,

I have this on my webpage:

PHP Code:

<?php

$selectnews = mysql_query("SELECT * FROM `news` ORDER BY `id` DESC LIMIT 3");

                while ($getnews = mysql_fetch_array($selectnews)){

                        echo "<li><h3>";
                        echo $getnews["date"];
                        echo "</h3><p>";
                        echo $getnews["message"];
                        echo "</p></li>";

                }

?>

And I keep getting this message:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a3674547/public_html/footer.php on line 45

I did some research on Google and I think it's having a problem connecting to the database, but I'm not 100% sure... This is how I structured the mysql_connect code:

PHP Code:

<?php
$dbhost = 'mysql3.000webhost.com';
$dbuser = 'a3674547_****';
$dbpass = '****';

$connect = mysql_connect($dbhost, $dbuser, $dbpass) or die(mysql_error());

$dbname = 'a3674547_****';
mysql_select_db($dbname) or die(mysql_error());

?>

Does anyone know what's wrong?

[flames]

change

PHP Code:

$dbhost = 'mysql3.000webhost.com'; 
and
$connect = mysql_connect($dbhost, $dbuser, $dbpass) or die(mysql_error()); 
and
mysql_select_db($dbname) or die(mysql_error()); 


PHP Code:

$dbhost = 'localhost'; 
and
$connect = mysql_connect($dbhost, $dbuser, $dbpass) );
and
mysql_select_db($dbname,$connect) or die(mysql_error()); 


[Devastator]

I would probably add a "or die(mysql_error())" to the mysql_query line. The fact that there are no errors during the connection and database selection tells me that there is nothing wrong with that section of the code.

In fact, I'm led to believe that the mysql_query is returning a null. With $selectnews being null, mysql_fetch_array complains because there's nothing there to fetch from.

[CKPD]

Flames, your code wouldnt work because its mot the localhost. The hos tis the 000webhost host.

Try this (Similar to what youve got just less complicated.

PHP Code:

<?php
   
  $host =  'mysql3.000webhost.com'; $user = 'YOUR USER'; $pass = 'YOUR PASS'; $db = 'DATABASE NAME';

  mysql_connect($host,$user,$pass)or die(mysql_error());
  mysql_select_db($db)or die(mysql_error());

?>

Rather then fiddling with all this other rubbish and making variables for ecerything, just keep it plain and simple and do the query without making it into a variable etc.

I would also change your query to:

PHP Code:

<?php

  $selectnews = mysql_query("SELECT * FROM `news` ORDER BY `id` DESC LIMIT 3");

                while ($getnews = mysql_fetch_assoc($selectnews)){

                        echo "<li><h3>".$getnews['date']."</h3>"
                        echo "<p>".$getnews['message']."</p></li>";

                }

?>

I have changed the "fetch_array" to "fetch_assoc" which does pretty much the same thing; only has a minor different which you can check on php.net. As has already been said you may have a null value in there but that WOULDN't return the error youve been given. It would just echo nothing, in other words give a blank space.

That error is when you have something wrong in your code for example:

PHP Code:

<?php

  $query = mysql_query("SELECT * FROM `news` WHERE `name` = '$name'");
  $result = mysql_fetch_array($yreuq);

 echo ''.$result['news'];

?>

Notice how i spelt query backwards. That is when you would get that sort of error because it cant find anything with that variable. If i were you id check all your spelling etc as ive made that mistake before!

[DJDrew22200]

Thanks for the tips guys, but nothing is working... Both of these codes are in the HEADER.PHP file:

PHP Code:

<?php
   
  $host =  'mysql3.000webhost.com'; $user = 'a3674547_user'; $pass = 'mypassword'; $db = 'a3674547_main';

  mysql_connect($host,$user,$pass)or die(mysql_error());
  mysql_select_db($db)or die(mysql_error());

?>

PHP Code:

<?php

$selectnews = mysql_query("SELECT * FROM `news` ORDER BY `id` DESC LIMIT 3");

    while ($row = mysql_fetch_assoc($selectnews)){

                echo '<ul>
                    <li>
                                <h3>'.$row["date"].'</h3>
                                   <p>'.$row["message"].'</p>
                    </li>
                </ul>';

    }

?>

There IS a table called "news" and it has 4 rows in it... I'm still getting this error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a3674547/public_html/header.php on line 74

Line 74 is:

while ($row = mysql_fetch_assoc($selectnews)){

Any other suggestions?

Thanks for all the tips so far guys!

[CKPD]

Check your private message!

[DJDrew22200]

Hey, I figured it out... I was asking it to sort by `id` by that column doesn't exist lol...

I meant `newsid`

[CKPD]

Nice one dude