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up vote 1 down vote favorite

I want to update the database with a checkbox in checked state.

  • If it is checked then update the database with 1.
  • Else if it is unchecked then update it with 0.

It works fine but it doesn't work if unchecked.

<?php 
include('lib/db.php');

$facebook_id ="10001088";
$query1 = "SELECT `video`,`quran`,`medical`,`groups`  FROM `man_facebook`.`users` WHERE `facebook_id`='$facebook_id'";
        $result1 = mysql_query($query1);
while($result = mysql_fetch_array($result1))
{
    $video = $result['video'];
    $quran = $result['quran'];
    $medical = $result['medical'];
    $groups = $result['groups'];
    echo $video;
 //   echo $quran;
?>

<form method="post"  action="<? echo $_SERVER['REQUEST_URI']; ?>" >

<input type="checkbox" name="video" id="video" value="<?echo $video;?>" <?php
if($video == '1'){
     echo "checked='checked'";
}
 else {}
echo "/>"
?> 
 <input type="submit" name="submit" value="Submit">
</form>

<?php
}
 if (isset($_POST['submit']))
{
 if (is_numeric($_POST['video']) && $_POST['video'] <2 )
{
 $video1 = isset($_POST['video']) ? '1' : '0';


 echo $video1;
    $query = mysql_query("UPDATE `man_facebook`.`users` 
SET `video` ='$video1'
WHERE `facebook_id`='$facebook_id'");

$video = $video1;
echo '<meta http-equiv="refresh" content="0" />';

    }
    }
//echo $query;
 //header("Location: updatesql.php"); 

?>

Can I also use jquery to update it smoothly?

share|improve this question
1  
Please do not use mysql_query in new applications. This is an interface from the 1990s that's in the process of being retired. You should be using mysqli and PDO to benefit from reliable, proper SQL injection protection. – tadman Aug 21 '12 at 6:18
U can and definitely should jquery to update the checkbox! – Saurabh Aug 21 '12 at 6:21
okay i will use mysqli but i want to solve that problem first – Salem Ahmed Aug 21 '12 at 6:40

1 Answer

up vote 0 down vote accepted

I find this part a bit weird:

if (is_numeric($_POST['video']) && $_POST['video'] <2 )
{
    $video1 = isset($_POST['video']) ? '1' : '0';

You first check if it is numeric and then check if it is set. If it wasn't set, that if would become false automatically. In other words, $video is always 1. Assuming $video can be only true/false (or maybe it comes as "checked"/"unchecked", not really sure), use it like this: 

$video1 = ($video ? '1' : '0');

Hopefully I spotted the issue successfully :)

UPDATE

<input type="checkbox" name="video" id="video" value="video" <?php
if($video == '1'){
     echo "checked='checked'";
}
echo "/>";

...

if (isset($_POST['submit']))
{
    echo $_POST['video']; // again, please tell what it outputs here!!!
    $video1 = (($_POST['video'] == "video") ? '1' : '0');
share|improve this answer
$video1 is always 0 now :) why i should don't use $_POST['video'] ??? – Salem Ahmed Aug 21 '12 at 6:29
No no, I mean you are checking it in some wrong way. Just echo $_POST['video'] in both cases(checked and unchecked) and tell me what you get :P – Andrius Naruševičius Aug 21 '12 at 6:30
i got 0 for ever :) – Salem Ahmed Aug 21 '12 at 6:32
Maybe it is because you also define it's value? Try getting rid of this part: ` value="<?echo $video;?>" ` – Andrius Naruševičius Aug 21 '12 at 6:35
Oh wait, actually it should be like this: updating the question... – Andrius Naruševičius Aug 21 '12 at 6:38
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