对于这种情况，阵列是否比交换机情况更有效？

Question

I'm displaying star ratings based on input from an API (note: only displaying, not receiving ratings at all). The following code works exactly as I need it to, but it's a very large piece of logic for something that feels like it could be massively simplified.

I read a few other tickets here that suggest that using an array might be more effective than a switch case. But because the logic depends on each case between true within a range of numbers (like more than 2.5 but less than 3) I'm not sure an array would work at all in this case.

Bottom line is this: Can this code be massively simplified somehow?

$stars = 3.5;

switch ($stars) {
  case ($stars > 1 && $stars <= 1.5):
    $star2 = 'dashicons-star-half';
    $star3 = 'dashicons-star-empty';
    $star4 = 'dashicons-star-empty';
    $star5 = 'dashicons-star-empty';
    break;
  case($stars > 1.5 && $stars <= 2):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-empty';
    $star4 = 'dashicons-star-empty';
    $star5 = 'dashicons-star-empty';
    break;
  case($stars > 2 && $stars <= 2.5):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-half';
    $star4 = 'dashicons-star-empty';
    $star5 = 'dashicons-star-empty';
    break;
  case($stars > 2.5 && $stars <= 3):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-filled';
    $star4 = 'dashicons-star-empty';
    $star5 = 'dashicons-star-empty';
    break;
  case($stars > 3 && $stars <= 3.5):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-filled';
    $star4 = 'dashicons-star-half';
    $star5 = 'dashicons-star-empty';
    break;
  case($stars > 3.5 && $stars <= 4):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-filled';
    $star4 = 'dashicons-star-filled';
    $star5 = 'dashicons-star-empty';
    break;
  case($stars > 4 && $stars <= 4.5):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-filled';
    $star4 = 'dashicons-star-filled';
    $star5 = 'dashicons-star-half';
    break;
  case($stars > 4.5):
    $star2 = 'dashicons-star-filled';
    $star3 = 'dashicons-star-filled';
    $star4 = 'dashicons-star-filled';
    $star5 = 'dashicons-star-filled';
    break;
  default:
  $star2 = 'dashicons-star-empty';
  $star3 = 'dashicons-star-empty';
  $star4 = 'dashicons-star-empty';
  $star5 = 'dashicons-star-empty';
}

?>
 <div class="wporg-ratings" title="<?php echo $stars; ?> out of 5 stars" style="color:#e6b800;">
   <span class="dashicons dashicons-star-filled"></span>
   <span class="dashicons <?php echo $star2; ?>"></span>
   <span class="dashicons <?php echo $star3; ?>"></span>
   <span class="dashicons <?php echo $star4; ?>"></span>
   <span class="dashicons <?php echo $star5; ?>"></span>
 </div>

Answer 1

I don't think an array or a switch are needed. You can use a for loop, and check the $stars value against your loop variable to see which icon should be used.

<div class="wporg-ratings" title="<?php echo $stars; ?> out of 5 stars" style="color:#e6b800;">
<?php
// for loop: one iteration for each of five possible stars
// (for loops are generally best for repeating code a specific number of times)
for ($i=0; $i < 5; $i++) {
    if ($stars <= $i ) {
        // empty stars are displayed when the iterator (i) is >= the star value
        echo  '<span class="dashicons dashicons-star-empty"></span>';
    } elseif ($stars <= $i + 0.5) {
        // half stars are displayed for star values between i and i+0.5 
        echo  '<span class="dashicons dashicons-star-half"></span>';
    } else {
        // whole stars are displayed when the star value is > i+0.5
        echo  '<span class="dashicons dashicons-star-filled"></span>';
    }
}
?>
</div>

To help understand why this works, take a theoretical value through the loop. We can use the one from your question, 3.5.

• first iteration: 3.5 > 0, 3.5 > 0.5, so you get the else value (filled star)
• second iteration: 3.5 > 1, 3.5 > 1.5, so you get the else value (filled star)
• third iteration: 3.5 > 2, 3.5 > 2.5, so you get the else value (filled star)
• fourth iteration: 3.5 > 3, 3.5 = 3.5, so you get the elseif value (half star)
• fifth iteration: 3.5 < 4, so you get the if value (empty star)