Warning: mysql_fetch_assoc(): Can someone help me with this?

Discussion in 'PHP' started by nuworldslack, Apr 13, 2012.

0
  1. #1
    Im not sure how to fix this problem i get the mesg: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a326878750/public_html/tree/database.php on line 255

    here is a piece of the code any help plz?
    
mysql_query("UPDATE videoer SET video_views=$views WHERE video_id=$id"); 
    }
    PHP:
    
    function video_alle($order) {
        if(!isset($_GET['side']))
        {
        $lala2 = "0";
        }
        else
        {
        $tal1 = $_GET['side'];
        $tal2 = $tal1 * 16;
        $tal3 = $tal2 - 16;
        $lala2 = $tal3;
        }
        $query = mysql_query("SELECT * FROM videoer ORDER BY $order DESC LIMIT $lala2, 16");
        while($row = mysql_fetch_assoc($query))
        {
            if($row['video_type'] == "youtube")
            {
            echo 
                '
                <li>
                    <div class="vediodiv">
                    ';
                    if($row['video_views'] >= "15000") { echo '<img style="position: absolute;" src="images/icon_pop_da.png" />'; }
                    if($row['video_id'] >= "2222" && $row['video_views'] <= "4999") { echo '<img style="position: absolute;" src="images/icon_ny_da.png" />'; }
                        echo '
                        <a href="video.php?id=' . $row['video_id'] . '">
                            <img width="147" height="99" src="http://img.youtube.com/vi/' . $row['video_url'] . '/0.jpg" 
                            
                        </a>
                   </div>
                   <div class="clear"></div>
                        <p class="vid-titel">
                            <a class="colr2"  href="video.php?id=' . $row['video_id'] . '">' . $row['video_titel'] . '
                            </a>
                        </p>
                            <p class="viewcounts">' . number_format($row['video_views']) . ' views</p>
                            <p class="by" >Posted byaf: <a href="#" class="undrlne">admin</a></p>
                   </li>
                   ';
            }
            if($row['video_type'] == "goviral")
            {
                echo 
                '
                <li>
                    <div class="vediodiv">
                    ';
                    if($row['video_views'] >= "15000") { echo '<img style="position: absolute;" src="images/icon_pop_da.png" />'; }
                    if($row['video_id'] >= "2222" && $row['video_views'] <= "4999") { echo '<img style="position: absolute;" src="images/icon_ny_da.png" />'; }
                        echo '
                        <a href="video.php?id=' . $row['video_id'] . '"><img src="http://pinoyspinays.com/reklamer/' . $row['video_tumb'] . '" alt="" />
                        </a>
                   </div>
                   <div class="clear"></div>
                        <p class="vid-titel">
                            <a class="colr2"  href="video.php?id=' . $row['video_id'] . '">' . $row['video_titel'] . '
                            </a>
                        </p>
                            <p class="viewcounts">' . number_format($row['video_views']) . ' views</p>
                            <p class="by" >Posted byaf: <a href="#" class="undrlne">admin</a></p>
                   </li>
                   ';
            }
        }
    }
    PHP:
     
    Last edited: Apr 13, 2012
    nuworldslack, Apr 13, 2012 IP
  2. sarahk

    sarahk iTamer Staff

    Messages:
    28,899
    Likes Received:
    4,555
    Best Answers:
    123
    Trophy Points:
    665
    #2
    The first thing to do is to change this 
            $tal1 = $_GET['side'];
        $tal2 = $tal1 * 16;
        $tal3 = $tal2 - 16;
        $lala2 = $tal3;
        }
        $query = mysql_query("SELECT * FROM videoer ORDER BY $order DESC LIMIT $lala2, 16");
    PHP:
    to this 
            $tal1 = intval($_GET['side']);
        $tal2 = $tal1 * 16;
        $tal3 = max($tal2 - 16,0);  //what if 'side' is missing or zero?
        $lala2 = $tal3;
        }
        $sql = "SELECT * FROM `videoer` ORDER BY {$order} DESC LIMIT {$lala2}, 16";
        echo "<hr>{$sql}</hr>";
        $query = mysql_query($sql) or die("Bad SQL: {$sql}<br>".mysql_error());
    PHP:
    Now, grab the sql that gets echo'd out and look at. Any obvious problems? Yes - fix them; No - copy the sql into phpMyAdmin, SqlYog, Toad or whatever you use to manage the database and run the query. You'll get a better error message from that if the query is at fault.
     
    sarahk, Apr 13, 2012 IP