title: use Img source with php variable?
body: <p>i have the following code in my detailansicht.php:</p>
  
  <pre><code>&lt;div class="file_description_box"&gt;
      &lt;b&gt;Beschreibung:&lt;/b&gt;&lt;br /&gt;&lt;br /&gt;
          &lt;?php 
               if(!empty($beschreibung))
                   echo '&lt;div align="center"&gt;';
                   echo '&lt;img src="$coverlink" alt="Cover"&gt;';
                   echo '&lt;/div&gt;&lt;br /&gt;';
                   echo format_content($beschreibung); 
               **else**
               echo '&lt;i&gt;Keine Beschreibung vorhanden&lt;/i&gt;';
          ?&gt;
  &lt;/div&gt;
  </code></pre>
  
  <p>but i think there has to be a mistake in the img tag. everytime i trie to open the page, he shows me an error: "Parse error: syntax error, unexpected T_ELSE in bla/bla/include/detailansicht.php on line 123"  (Line 123 is the else marked bold).
  I tried several methods but i always get this error.
  It would be nice, if someone could help me with this.</p>
  
  <p>-sTarbuZz</p>