Compare and show different data from two tables

Question

I've got two tables. Skills, that contains all the available skills. and user_skills, that contains which skills user got. I need to output all the available skills, and those skills that the user have (I select it by session['username'] ) will be marked (checked checkbox).

$get_all_skills = mysqli_query($conn, "SELECT * FROM skills");
$web_design = array();
$web_develop = array();
$automation = array();
$security = array();
while($show_row = mysqli_fetch_array($get_all_skills)){
    switch($show_row['skill_type']){
        case '1':
            array_push($web_design, $show_row['skill_name']);
            break;
        case '4':
            array_push($web_develop, $show_row['skill_name']);
            break;
        case '3':
            array_push($automation, $show_row['skill_name']);
            break;
        case '2':
            array_push($security, $show_row['skill_name']);
            break;
    }
}

How can possibly I make this work? Part of the html is:

<div class="">
                                <ul class="to_do">
                                  <?php
                                  for($i=0;$i<count($web_develop);$i++){

                                  ?>
                                  <li>
                                    <p>
                                      <input type="checkbox" class="flat"> <?php echo $web_develop[$i];?> </p>
                                  </li>
                                  <?php } ?>
                                </ul>
                              </div>

Answer 1

Considering you have a user_skills array ($user_skills_array) declared and have the data, use the below html code

<div class="">
  <ul class="to_do">
  <?php
    for($i=0;$i<count($web_develop);$i++){
      // check the global skill is available in the user skill set
      $checked = (in_array($web_develop[$i], $user_skills_array)) ? 'checked="checked"': ''; 
   ?>
    <li>
      <p>
        <input type="checkbox" class="flat" <?php echo $checked;?>> <?php echo $web_develop[$i];?> </p>
    </li>
    <?php } ?>
  </ul>
</div>

Answer 2

try the following, explanation below.

 <?php
$get_all_skills = mysqli_query($conn, "SELECT * FROM skills");

$availableSkills = [];
$skillTypes = [
    1 => 'Webdesign',
    2 => 'Security',
    3 => 'Automation',
    4 => 'Web Development',
];

while($show_row = mysqli_fetch_array($get_all_skills)){
    $skill = [
        'id' => $show_row['id'];
        'name' => $show_row['name'];
    ];
    $skillType = $show_row['skill_type'];

    $availableSkills[$skillType][] = $skill;
}

// assuming user_skills:
$userSkills = [
    1,
    2,
    5,
    8
];

?>

    <ul class="to_do">
      <?php foreach ($availableSkills as $type => $skillsForType) : ?>
        <li>
            <?= $skillTypes[$type] ?>
            <ul>
                <?php foreach ($skillsForType as $skill) : ?>
                    <li>
                        <input 
                            type="checkbox" 
                            class="flat" 
                            name="skill[<?= $type ?>]" 
                            value="<?= $skill['id'] ?>"
                            <?= in_array($skill['id'], $userSkills) ? 'checked="checked"' : '' 
                        > <?= $skill['name'] ?>
                    </li>
                <?php endforeach; ?>
            </ul>
        </li>   
      <?php endforeach; ?>
    </ul>

First you gather all skills, as have you already done. Next I optimized your different skills array into one and listed all skill types.

The available skills are indexed by their type id (see $skillTypes).

Next I assumed your user-skill collection could be like shown.

In the HTML section I am iterating over every skill type and in a nested foreach echo all corresponding skills. If the skill id of the current iteration is in_array of the user-skill-relation the "checked" attribute is set.

To clarify I used short syntax in this example.

[] --> array()
<?= --> <?php echo

foreach () : endforeach;