Check a checkbox when element exists in the database - php

Question

I have a table that has different names of competences and another one in which we have the competence_ID and the user_ID that has that competence. I have a form in which I submit the competences that apply to a certain user. Now, I want to retrieve these competences and check boxes in case they are valid.
I use this code, but it does not seem to work.
<?php
include "db.php";
$employees=mysql_query("SELECT * FROM asiakas");
echo "<form method='post' action='' id='employeesselection'><select name='select_employee' id='select_employee'>";
while($row=mysql_fetch_array($employees)){
$selected = ($row['Id'] == $_POST['select_employee'])?'selected="selected"':'';
echo '<option '.$selected.' value="'.$row['Id'].'">'.$row['Etunimi'].' - '. $row['Sukunimi'].'</option>';
}
echo "</select><input type='hidden' name='action' value='selectedemployee'>
<input type='submit' value ='submit' name='submit'><br/>";
if(isset($_POST['select_employee'])){
$specific=mysql_query("SELECT Id, Sukunimi,Etunimi from asiakas WHERE Id=".$_POST['select_employee']);
$sp=mysql_fetch_array($specific);
}
if(isset($sp['Etunimi']) && isset($sp['Sukunimi'])){
$comp=mysql_query("SELECT * FROM Competences");
echo "<br/>select competences for: ". $sp['Etunimi'];
$id= $sp['Id'];
$result=mysql_query("SELECT distinct c_ID from User_Competence WHERE e_ID=".$id);
while($test=mysql_fetch_array($result))
{
echo $test['c_ID'];
}
echo "<table><th>valid?</th><th>Competence description</th>";
$counter=0;
$traverse=0;
while($compi=mysql_fetch_array($comp)){
$checked='';
if($test[$traverse]==$counter){
$checked="checked";
$traverse++;
}
$counter ++;
echo "<tr><td><input type='checkbox'" .$checked." name='c[]' value='".$compi['Competence_ID']."'></td><td>".$compi['Competence_Description']."</td></tr>";
}
echo "</table>";
echo "<input type='hidden' name='action' value='selectchecked'>";
echo "<input type='submit' value='submit checks'>";
}
if(isset($_POST) && !empty($_POST)){
print_r($_POST);
}
if(isset($_POST['action']) && $_POST['action']='selectchecked'){
if (isset($_POST['c'])){
$s = $_POST['c'];
foreach ($s as $k => $v) {
if (is_array($v)) {
// array_push($s, implode('', array($v [$what])));
}
};
echo $abc= implode(',', $s);
for ( $a=0;$a<count($s);$a++){
$ar=explode(',',$abc);
echo $var= $ar[$a];
$q=mysql_query("INSERT INTO user_competence(c_ID, e_ID) VALUES ('".$var."','".$id."')");
}
}
}
echo "</form>";
?>

Answer 1

If you put checked as an attribute on the <input type="checkbox"> it will be checked, even if you leave the actual value of the attribute empty.
So change the following parts from...
$checked="checked";
...
echo "<tr><td><input type='checkbox' checked='".$checked."' name='c[]' ...
Into...
$checked="checked='checked'";
...
echo "<tr><td><input type='checkbox' ".$checked." name='c[]' ...
Which will mean if the $test[$traverse$] is true, checked='checked' will be placed into the HTML, otherwise it will be left out.