Set mysql values from radio buttons in html form

Question

I want two radio buttons on a webpage (written in php) representing "yes" and "no". When I load the page I want it to fetch a value from a mysql db and set the corresponding radio button. And when I click on the other button, I want it to update the database and reload the page.

I'm trying to do this with a simple html form, but no luck. The code I have so far (that is not working at all :( is:



if (!isset($_POST['submit'])) {
    $sql = "SELECT challenge_me FROM contestants WHERE id=$id";
    $res = (mysql_fetch_assoc(mysql_query($sql, $db)));
    $challenge_me = $res["challenge_me"];

}else{
    $sql = "UPDATE contestants SET challenge_me='" . $_POST['YesNo'] . "' WHERE id='$id'";
    if(!mysql_query($sql, $db))
        echo mysql_error(), "<br/>Query '$sql'";
    $challenge_me = $_POST['YesNo'];
}

    echo'<form method="post" action="' . $PHP_SELF . '">';
    echo '<input type="hidden" name="submit" value="submit">';          
if($challenge_me == 1){
    echo'<input type="radio" name="YesNo" value="1" onClick="this.form.submit();" checked>Yes ';
    echo'<input type="radio" name="YesNo" value="0" onClick="this.form.submit();">No ';
}else{
    echo'<input type="radio" name="YesNo" value="1" onClick="this.form.submit();">Yes ';
    echo'<input type="radio" name="YesNo" value="0" onClick="this.form.submit();" checked>No ';
}
echo'</form>';

Not really an answer to your question, but you shouldn't be using string concatenation to build an SQL query. Your script is currently vulnerable to SQL injection based on using $_POST['YesNo'] without at least escaping it or using mysql statements. — Rob
– Rob, Commented May 10, 2011 at 22:09
Hmm, seems like I solved it. It didn't work if the hidden post was named "submit". That prevented the form to do anything when I clicked on the radio buttons. Renaming it to something else worked well... And thanks for the comment about sql injection! — faximan
– faximan, Commented May 11, 2011 at 5:41

boug · Accepted Answer · 2011-05-10 22:43:10Z

1

Your script doesnt seem to define $id, where does $id get its value from? That could be the source of your problem. Your script might not be passing any value in $id

answered May 10, 2011 at 22:43

boug

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4 Comments

dqhendricks Over a year ago

also, $PHP_SELF is blank i'm assuming. I think you are trying to use $_SERVER['PHP_SELF']. this probably isn't affe4cting the code however because leaving the action blank will automatically submit the form to itself.

faximan Over a year ago

Well, $id comes as an argument to the function (I removed a couple of lines of code that I thought wasn't interesting for my question).

boug Over a year ago

@faximan have you resolved the problem yet? have you tried @dqhendricks suggestion, that is another possible cause

faximan Over a year ago

Yeah, thanks. It works now, all that was needed was to change the name of the hidden input from "submit" to "submitValue".Guess "submit" is some kind of reserved word.

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