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Home/ Questions/Q 7520693
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T02:06:46+00:00 2026-05-30T02:06:46+00:00

Is there a better way to express this PHP code? I’ve received variables from

  • 0

Is there a better way to express this PHP code? I’ve received variables from a form and now need to check whether they are null or not before outputting a text string.

Many thanks in advance!

$Capacity = $_POST["Capacity"];
$Location = $_POST["Location"];
$RoomType = $_POST["RoomType"];

echo '<div id="container" style="padding: 15px; background-color: #949494;">';
echo '<div style="padding: 15px; background-color: #fff;">';
echo '<h2>Search results:</h2>';
echo '<h3>Showing';
if ($RoomType == '') echo ''; else echo ' '.$RoomType.' ';
echo ' rooms with a capacity of '. $Capacity .' and over';
if ($Location == '') echo ''; else echo ' in  '.$Location.' Park';
echo '</h3>';
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T02:06:47+00:00Added an answer on May 30, 2026 at 2:06 am

    Edit 2: I revisited the code, so that it makes more sense.

    <?php
    $requiredFields = array(
        'Capacity',
        'Location',
        'RoomType'
    );
    // Retrieve only the allowed fields
    $fields = array_intersect_key($_POST, array_flip($requiredFields));
    // Remove whitespace
    $fields = array_map('trim', $fields);
    // Remove empty values
    $fields = array_filter($fields, 'strlen');
    // Filter values for HTML output
    $fields = array_map('htmlentities', $fields);
    
    // We require all fields
    if(count($fields) !== count($requiredFields)) {
        exit('Your browser sent incorrect data!');
    }
    
    // Create the variables
    extract($fields, EXTR_SKIP);
    ?>
    <h2>Search Results</h2>
    <h3><?php printf('Showing %s rooms with a capacity of %s and over in %s Park', $RoomType, $Capacity, $Location) ?></h3>
    

    Edit: Added htmlentities filter

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