how to display multiple rows from sql database into a checkbox based off same foreign key

Question

I have a questionnaire that has a list of check-boxes for ages that save into database into each differnt rows but same foreign id (see image below the select statement as well as the results i am getting). Now I want to take those 4 results and put them back into a checkbox list where the checkboxes if there is a value is checked, the way I got it to work in the actual questionnaire (if say the validation failed was through the querystring so the code looked like code slot 1 below)

but now I am using a CMS so it is taking the results straight from the database so im not so sure how to get the 4 values into an array i can use to check in_array for the code slot shown below.

PHP Code from the questionaire.

<?php
  if (isset($_GET['kage']) && $_GET['kage'] !== null) {
  $agegroups = explode("_", $_GET['kage']);

  $checked = in_array('kids', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="kids" id="kid"><input type="checkbox" name="age[]" class="ages" value="kids" id="kids" ' . $checked . '>Ages 0 - 10</label>';
  $checked = in_array('teens', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="teens" id="teen"><input type="checkbox" name="age[]" class="ages" id="teens" value="teens" ' . $checked . '> Ages 11-19 </label>';
  $checked = in_array('twenties', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="twentys" id="twenty"><input type="checkbox" name="age[]" class="ages" id="twentys" value="twenties" ' . $checked . '> Ages 20-29</label>';
  $checked = in_array('thirties', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="thirtys" id="thirty"><input type="checkbox" name="age[]" class="ages" id="thirtys" value="thirties" ' . $checked . '> Ages 30-39 </label>';
  $checked = in_array('forties', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="fourtys" id="fourty"><input type="checkbox" name="age[]" class="ages" id="fourtys" value="forties" ' . $checked . '> Ages 40-49</label>';
  $checked = in_array('fifties', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="fiftys" id="fifty"><input type="checkbox" name="age[]" class="ages" id="fiftys" value="fifties" ' . $checked . '> Ages 50-59 </label>';
  $checked = in_array('sixtyup', $agegroups) ? 'checked="checked"' : '';
     echo '<label for="sixtyplus" id="sixty"><input type="checkbox" name="age[]" class="ages" id="sixtyplus" value="sixtyup" ' . $checked . '> Ages 60+</label>';
  echo '<br>';
} else {
 echo '<label for="kids" id="kid"><input type="checkbox" name="age[]" class="ages" id="kids" value="kids">Ages 0 - 10</label>  
       <label for="teens" id="teen"><input type="checkbox" name="age[]" class="ages" id="teens" value="teens"> Ages 11-19 </label>
       <label for="twentys" id="twenty"><input type="checkbox" name="age[]" class="ages" id="twentys" value="twenties"> Ages 20-29</label>
       <label for="thirtys" id="thirty"><input type="checkbox" name="age[]" class="ages" id="thirtys" value="thirties"> Ages 30-39 </label>
       <label for="fourtys" id="fourty"><input type="checkbox" name="age[]" class="ages" id="fourtys" value="forties"> Ages 40-49</label>
       <label for="fiftys" id="fifty"><input type="checkbox" name="age[]" class="ages" id="fiftys" value="fifties"> Ages 50-59 </label>
       <label for="sixtyplus" id="sixty"><input type="checkbox" name="age[]" class="ages" id="sixtyplus" value="sixtyup"> Ages 60+</label>';
}
?>

When i try using

//where $vId = 2;
$sql_elite_ages = "select * from inf_ages_interacted where inf_id = $vId";
$rs_elite_ages = mysqli_query($vconncvnl, $sql_elite_ages);
$rs_elite_ages_all = mysqli_fetch_assoc($rs_elite_ages);
$rs_elite_ages_array = mysqli_fetch_array($rs_elite_ages); 

combined with

$agegroups = $rs_elite_ages_array;
$array= ' ';

foreach ($agegroups as $row){
echo $addary .= $row .' ';
}

my results are "2 2 2 2 2 2 2 2 2 2 2 2 2 2 teens 2 2 2 2 teens teens"

what i want to show is

so the age group values match the relevant valued checkbox and if the value is present then the box is checked otherwise leave the box unchecked

so when i do this way @Lelio Faieta

$sql="SELECT `agegroup` FROM `inf_ages_interacted` WHERE inf_id = $vId";
$rs = mysqli_query($vconncvnl,$sql);

while($row = mysqli_fetch_assoc($rs)){

print_r($row); echo '<br>';

} the results i get is 

Answer 1

Instead of use foreach() use while()

This is the intended functionality of mysql_fetch_array(). If you want to not have the "duplicates" and just have the associative-array, use mysql_fetch_assoc() instead.

Example:

while ($row = mysql_fetch_assoc($data)) { ... }