Keep checkbox checked, after refresh and submit

Question

I'm trying to keep a checkbox checked, but after I refresh the page, if the checkbox is checked the database will get updated with 'yes', but the database automatic updates it to 'no' even if the checkbox is checked with local.store, I hope somebody can help me. This is my code.

    <form action="" method="POST">
     <label for="option1">Waarschuwingsbericht inschakelen voordat het volgende pack wordt geopend als jou pack boven de 200.000 waard is?</label><input id='option1' type="checkbox" name="checkbox" value="yes"><br>
      <input type="submit" value="Opslaan en nog een pack openen">
    </form>
<?
    if(isset($_POST['checkbox'])){
        $sql = "UPDATE users SET puntenchecked = 'yes' WHERE username = '" . $usernamez . "'";
        $result = mysql_query($sql) or die('Query failed: ' . mysql_error());
    }
    else {
        $sql = "UPDATE users SET puntenchecked = 'no' WHERE username = '" . $usernamez . "'";
        $result = mysql_query($sql) or die('Query failed: ' . mysql_error());
    }
?>
<script>
$(function(){
    var test = localStorage.input === 'true'? true: false;
    $('input').prop('checked', test || false);
});

$('input').on('change', function() {
    localStorage.input = $(this).is(':checked');
    console.log($(this).is(':checked'));
});
</script>

You dont need this $('input').prop('checked', test || false);. This should be enough $('input').prop('checked', test); — Rajesh Dixit
– Rajesh Dixit, Commented Jan 30, 2016 at 10:04
Use checked attribute into <input type="checkbox"> based on $_POST['checkbox'] request is set. See here stack answer already did. — AddWeb Solution Pvt Ltd
– AddWeb Solution Pvt Ltd, Commented Jan 30, 2016 at 10:06

Amit Visodiya · Accepted Answer · 2016-01-30 10:03:13Z

2

try this :

<input id='option1' type="checkbox" name="checkbox" value="yes" <?php echo (isset($_POST['checkbox']))? "checked='checked'": "";?> >

answered Jan 30, 2016 at 10:03

Amit Visodiya

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AddWeb Solution Pvt Ltd · Accepted Answer · 2016-01-30 10:09:14Z

1

Replace below JS logic:

<script>
$(function(){
    var test = localStorage.input === 'true'? true: false;
    $('input').prop('checked', test);
});

$('input').on('change', function() {
    localStorage.input = $(this).is(':checked');
    console.log($(this).is(':checked'));
});
</script>

Check for more here.

answered Jan 30, 2016 at 10:09

AddWeb Solution Pvt Ltd

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