php - Debug MYSQLI Conversion -

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i'm still bit new @ mysqli please go easy. i'm trying convert mysqli , i'm coming with.

db_conx.php

<?php $db_conx = mysqli_connect("localhost", "use", "pass", "db"); // evaluate connection if (mysqli_connect_errno()) {     echo mysqli_connect_error();     exit(); } ?>  <?php     // connect mysql database      include "includes/db_conx.php";  $sql = "select * content order id desc";  $result = mysqli_query($db_conx,$sql); $productcount = mysqli_num_rows($result);          $bloglist = "";     if ($productcount > 0) {  $adverts = array( 'test', 'test 2'); $counter = 0;  while($row = mysqli_fetch_assoc($sqltwo)){ $id = $row["id"]; $article_title = $row["article_title"]; $category = $row["category"]; $author = $row["author"]; $date_added = $row["date_added"]; $article = $row["article"]; $short = substr(strip_tags($article), 0, 750); $shorttitle = substr(strip_tags($article_title), 0, 45);  if($readmore == ''){ //code new post $bloglist .= '<div class="blogsnippettitle"><a href="http://www..php?id='.$id.'"><h2>'.$article_title.'</h2></a></div><div class="blogsnippet"><div class="blogimage"><img src="http://www./'.$id.'.jpg" height="142px" width="200px" alt="'.$category.' '.$shorttitle.'" /></div><div class="blogsnippetprev"><div class="citation">by <span style="color:#006699;">'.$author.'</span> on <span style="color:#99aacc;">'.$date_added.'</span> in <span style="color:#006699;">'.$category.'</span></div> <div class="snippet">'.$short.'...<br /></div> <div class="readmorebutton"><br /><a href="http://www..php?id='.$id.'"><img src="http://www.read_more.png" alt="read more graphic" /></a></div> </div> </div>'; } else{ //code old post $bloglist .= '<div class="blogsnippettitle"><a href="'.$readmore.'"><h2>'.$article_title.'</h2></a></div><div class="blogsnippet"><div class="blogimage"><img src="http://www./'.$id.'.jpg" height="142px" width="200px" alt="'.$category.' '.$shorttitle.'" /></div><div class="blogsnippetprev"><div class="citation">by <span style="color:#006699;">'.$author.'</span> on <span style="color:#99aacc;">'.$date_added.'</span> in <span style="color:#006699;">'.$category.'</span></div> <div class="snippet">'.$short.'...<br /></div> <div class="readmorebutton"><br /><a href="'.$readmore.'"><img src="http://www.read_more.png" alt="read more graphic" /></a></div> </div> </div>'; } if($counter < 2){ $bloglist .= $adverts[$counter]; $counter++; } } } ?>

the error i'm getting

[07-aug-2013 22:18:26 america/denver] php warning: mysqli_fetch_assoc() expects parameter 1 mysqli_result, null given in /home/learnsit/public_html/site-design-blog.php on line 23

you passing sqltwo in mysqli_fetch_assoc() but takes argument of mysqli_result $result

try this,

while($row = mysqli_fetch_assoc($result)){

in place of 

while($row = mysqli_fetch_assoc($sqltwo)){

read mysqli_fetch_assoc()


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