John Davidson

Message:

Currently I am fetching all the values from my database and placing a checkbox next to them. By default all checkboxes are unchecked, but the checkboxes whose id is present in the database should be marked as checked. To do this I have used the following code:

View Class:

    <?php if($facilities) foreach($facilities as $facility):  ?>

     <?php foreach($checked_facility as $key => $value){ ?>

        <fieldset id="availablenetworked">

            <input type="checkbox" <?php echo ($value['facilities_id'] == $facility['id'] ? 'checked' : ''); ?> 

            name="facility[]" id="<?php echo $facility['id'] ?>" value="<?php echo $facility['id'] ?>"> 

            <label for="<?php echo $facility['id'] ?>"><?php echo $facility['title'] ?></label>

        </fieldset>

    <?php } endforeach; ?>

Doing this correctly checkmarks all the values that I have in my database, but messes up my view and makes it look like this:

$facilities is used to fetch all records of facilities and $checked_facility is used to check the values that a specific person has chosen.

The response returned from $checked_facility looks like this where I'm trying to access the facilities_id and marking all those with a checkmark in my list.

---

My Answer:
php
class View {
public function displayCheckboxes($valuesFromDatabase, $allValues) {
foreach ($allValues as $value) {
$checked = in_array($value['id'], $valuesFromDatabase) ? 'checked' : '';
echo '' . $value['name'] . '
';
}
}
}
?>

In the above code, the `displayCheckboxes` method takes two parameters: `$valuesFromDatabase` which is an array of ids present in the database, and `$allValues` which is an array of all values fetched from the database. It then loops through all values and checks if the id is present in the `$valuesFromDatabase` array. If it is present, the checkbox is marked as checked.