PHP - checkbox value / db insert and if(isset())

Question

I'm new to PHP, and I have stumble on the problem which I don't know how to solve. I'm 99% it is due my poor knowledge of PHP ( I'm PHP user since last Monday:) )

Just in front I will declarate that:

• db conncetion is working
• table does exist
• values are saved correctly to the db

I have following form:

<form id="loginForm" name="loginForm" method="post" action="../exe/news-exec.php">
      <input name="live" type="checkbox" class="textfield" id="live" />
      <input name="content" type="text" class="textfield" id="content" />
      <input type="submit" name="Submit" value="Register" />
</form>

And following file is executing this:

<?php
    //Start session
    session_start();

    //Include database connection details
    require_once('../inc/config.php');

    //Connect to mysql server
    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    if(!$link) {
        die('Failed to connect to server: ' . mysql_error());
    }

    //Select database
    $db = mysql_select_db(DB_DATABASE);
    if(!$db) {
        die("Unable to select database");
    }

    //Function to sanitize values received from the form. Prevents SQL injection
    function clean($str) {
        $str = @trim($str);
        if(get_magic_quotes_gpc()) {
            $str = stripslashes($str);
        }
        return mysql_real_escape_string($str);
    }

    //Sanitize the POST values
    $live = clean($_POST['live']);
    $content = clean($_POST['content']);



    if(isset($live)) { $live = 1;}
    if(!isset($live)) { $live = 0;}



    //Create INSERT query
    $qry = "INSERT INTO news(live, content) VALUES('$live','$content') ";
    $result = @mysql_query($qry);

    //Check whether the query was successful or not
    if($result) {
        echo $live."<br /><br />";
        echo '<a href="../">Index File</a>';

        exit();
    }else {
        die("Query failed");
    }
?>

What the form should do:

• if the checkbox is checked - save the value of '1' into field 'live' in the table 'news'
• if the checkbox is NOT checked - save the value of '0'

If the checkbox has been checked everything is working fine, but if the checkbox is not checked (should echo $live = 0 ), but is displaying value = 1 and following notice: Notice: Undefined index: live in C:\wamp\www\exe\news-exec.php on line 30

Line 30: $live = clean($_POST['live']);

I'm 99% sure the problem are those declaration:

if(isset($live)) { $live = 1;}

if(!isset($live)) { $live = 0;}

What I'm doing wrong? Any suggestion much appreciated.

Answer 1

HTML:

<input type="hidden" name="live" class="textfield" id="live0" value="0" /> 
<input type="checkbox" name="live" class="textfield" id="live1" value="1" />

PHP:

$live = clean($_POST['live']);

What happens here is that when the checkbox is left unchecked, the hidden field’s value gets submitted, as-is. When the check box is checked, the hidden field’s POST value gets overwritten by the activated checkbox’s.

Hope this helps.

Answer 2

Try this:

if (isset($_POST['live'])) $live=1; else $live=0;

Line 30: $live = clean($_POST['live']); causes isset($live) to be true, no matter if $_POST['live'] is set or not, so you have to check $_POST['live'] directly.