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PHP

Ahmed Lafeer
Ahmed Lafeer
20,181 Points
Posted by Ahmed Lafeer
Ahmed Lafeer
20,181 Points

Code Challenge: Accounting for Empty Results (2 of 3)

Can't seem to figure out why the code in the else block isn't working.

"Bummer! The array is empty, but I don't see the message indicating that no flavors are found."

<?php

    $recommendations = array();

?><html>
<body>

    <h1>Flavor Recommendations</h1>


      <?php foreach($recommendations as $flavor) { 
          if(!empty($recommendations)) { 
              echo "<ul>";
              echo "<li>" . $flavor . "</li>";
              echo "</ul>";
          } else {
          echo "<p>There are no flavor recommendations for you.</p>"; 
          }
      }
      ?>

</body>
</html>

3 Answers

Jason Anello
Jason Anello
Courses Plus Student 94,610 Points

Hi Ahmed,

You have a few problems with the code.

First, by bringing the ul into the foreach each flavor becomes it's own unordered list now.

Second, your foreach needs to be contained inside the if block. If the array is empty then the foreach won't execute at all and neither your if or else will be executed.

<h1>Flavor Recommendations</h1>

  <?php if (!empty($recommendations)) { ?>
    <ul>
      <?php foreach($recommendations as $flavor) { ?>
        <li><?php echo $flavor; ?></li>
      <?php } ?>
    </ul>
  <?php } else { ?>
    <p>There are no flavor recommendations for you.</p>
  <?php } ?>
Andreas cormack
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points
Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

Hi Ahmed

I had this same problem aswell I think .

wrap the if statement around the h1 see below

<?php if(!empty($recommendations)){

echo "<h1>Flavor Recommendations</h1>";
 foreach($recommendations as $flavor) { 
          echo "<ul>";
          echo "<li>" . $flavor . "</li>";
          echo "</ul>";
      }

} else { echo "<p>There are no flavor recommendations for you.</p>"; } ?>

try this .

Ahmed Lafeer
Ahmed Lafeer
20,181 Points
Ahmed Lafeer
Ahmed Lafeer
20,181 Points

Hey Andreas

Unfortunately that code doesn't seem to work either :/

I managed to complete the task by inserting a dummy element into the array as such, but I don't think that's the right approach. 

<?php    $recommendations = array("element"); ?>

Thanks for your help though!