PHP error problem

I get the following error on line 8: Undefined index: privacy_policy which is $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); I was wondering how can I fix this problem?

Here is line 8.

$privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']);

Here is the PHP.

if (isset($_POST['submitted'])) {

$mysqli = mysqli_connect("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT users.*
                             FROM users 
                             WHERE user_id=3");

$privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']);

    if (mysqli_num_rows($dbc) == 0) {
            $mysqli = mysqli_connect("localhost", "root", "", "sitename");
            $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, privacy_policy) 
                                         VALUES ('$user_id', '$privacy_policy')");
    }


    if ($dbc == TRUE) {
            $dbc = mysqli_query($mysqli,"UPDATE users 
                                         SET privacy_policy = '$privacy_policy' 
                                         WHERE user_id = '$user_id'");

 开发者_开发技巧           echo '<p class="changes-saved">Your changes have been saved!</p>';

    }


    if (!$dbc) {
            print mysqli_error($mysqli);
            return;
    }

}

Here is the HTML.

<form method="post" action="index.php">
    <fieldset>
        <ul>
            <li><input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" <?php if (isset($_POST['privacy_policy'])) { echo 'checked="checked"'; } else if($privacy_policy == "yes") { echo 'checked="checked"'; } ?> /></li>

            <li><input type="submit" name="submit" value="Save Changes" class="save-button" />
                <input type="hidden" name="submitted" value="true" />
            <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li>
            </ul>
    </fieldset>
</form>

If the warning is saying that $_POST['privacy_policy'] is undefined then that is (almost) undoubtedly true.

Your privacy policy value comes from a check box and if the check box is unchecked when you submit the form, then there will be no entry for it in the $_POST array. If it is checked when you submit the form then the value in $_POST['privacy_policy'] will be yes.

Before you assign the value in $_POST['privacy_policy'] to something, check that it is set with isset. This will stop your warning from appearing.

update

I would deal with an unchecked check box like this:

$privacy_policy = "no";
if(isset($_POST['privacy_policy']))
{
    $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']);
}

Add a hidden field with the default value if the "privacy_policy" checkbox is not selected. Then the $_POST variable will contain a value. When the checkbox is selected, it will override the value of the hidden field (as long as it follows the hidden field).

<form method="post" action="index.php">
<!-- add a default value for the privacy_policy -->
<input type="hidden" name="privacy_policy" value="no"/>
    <fieldset>
        <ul>
            <li><input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" <?php if($privacy_policy == "yes") { echo 'checked="checked"'; } ?> /></li>
            <li><input type="submit" name="submit" value="Save Changes" class="save-button" />
                <input type="hidden" name="submitted" value="true" />
            <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li>
            </ul>
    </fieldset>
</form>

EDIT: As Matt Ellen also correctly points out, you can check to see that $_POST['privacy_policy'] is set using isset() or array_key_exists() and set a default in the PHP as well.

$privacy_policy = array_key_exists('privacy_policy', $_POST) ? $_POST['privacy_policy'] : false;