Question

I want to fetch data from my MySQL and echo it out only if $featured == 1.
It works, the only one with $featured == 1 will show title, but my problem is that it also makes posts from everything else in my database (empty posts). I only want it to show the posts with == 1 and nothing more nothing less. I can’t manage to fix it.

<aside id="featured" class="body"><article> 
<?php
for ($i=0; $i <$num_results; $i++){
$row1 = $result1->fetch_assoc();
$featured = ($row1['featured']);
if ($featured == 1) {
echo $row1['title'];    
} else {
}
?>
<figure> 
<img src="images/black2.gif" alt="Black 2" style="width: 300px;"/> 
</figure>   
<hgroup>    
<h2>Featured Article</h2> 
<h3><a href="goodies/black2.html">
</a></h3> 
</hgroup>   
<p> </p> 
<footer class="post-info"> 
<abbr class="published" title="date">20 juli 2012</abbr>    
<address class="vcard author">By<a class="url fn" href="portfolio.html">F4LLCON</a> 
</address></footer><!-- /.post-info -->
</article>
<?php
}
$result1->free();   
?>

</aside><!-- /#featured -->

---

• Will look like:

---

---

• But I want it like this:

---

Answer 1

Your for body contains printing large amount of HTML regardless of featured checking, that is, outside your if.

You probably should move two lines

} else { 
}

line to the bottom of the code snippet.