PHP checkbox

[RTS]

I am working on a PM system for my site, and having trouble with a line of code for a checkbox. I want it so that if the ncheckbox is checked, it inserts the word "yes" in to the column "sig" in my mysql database. hers the code I have so far. It gives me [code]Parse error: parse error in /Library/WebServer/Documents/users/send.php on line 8[/code]

send.php:
[code]<?php
session_start();
$con = mysql_connect("localhost","ZackBabtkis","");
$subject = mysql_real_escape_string($_POST[subject]);
$message = mysql_real_escape_string($_POST[message]);
$user = $_COOKIE["user"];
$sender = $_SESSION['username'];
$date = date("F j, Y, g:i a");
$sig = if (isset($_POST["sig"])) echo "yes";
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("test", $con);
$sql = "INSERT INTO messages (Username, Sender, Subject, Message, Date, sig) VALUES ('$user','$sender','$subject','$message', '$date', '$sig')";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Your message has been sent to $user";
mysql_close($con)
?>[/code]

[.josh]

$sig = (isset($_POST["sig"])) ? "yes" : NULL;

[RTS]

okay, but how do I make it so that when the box isnt checked it insterts "no" in to the database?

[.josh]

change the NULL to "no"

$sig = (isset($_POST["sig"])) ? "yes" : "no";

or else in your database you can set the default to "no".

[Daniel0]

Small modification (looks nicer and is smaller):

[code]<?php
session_start();
$con = mysql_connect("localhost","ZackBabtkis","") or die('Could not connect: ' . mysql_error());
mysql_select_db("test", $con);

$subject = mysql_real_escape_string($_POST['subject']);
$message = mysql_real_escape_string($_POST['message']);

$date = date("F j, Y, g:i a");

$sig = isset($_POST['sig']) ? "yes" : "no";

mysql_query("INSERT INTO messages (Username, Sender, Subject, Message, Date, sig) VALUES ('{$_COOKIE['user']}','{$_SESSION['username']}','{$subject}','{$message}', '{$date}', '{$sig}')",$con)) or die('Error: ' . mysql_error());

echo "Your message has been sent to {$user}";

mysql_close($con)
?>[/code]

[RTS]

when I have [code]$sig = (isset($_POST["sig"])) ? "yes" : "no";[/code] it only echos no, when I change it to [code]$sig = (isset($_POST["sig"])) ? "no" : "yes";[/code] it always inserts yes. it seems to be ignoring the first quotes

[Daniel0]

Make sure that your checkbox has a non-empty value attribute.

[.josh]

then the problem is with your post variable itself

$sig = (isset($_POST["sig"])) ? "yes" : "no";

basically what this does is if $_POST['sig'] is set (it exists) then $sig = "yes";  if it does not exist, then $sig = "no";

the fact that it keeps setting $sig to the one on the right of the : means taht $_POST['sig'] is not set.  check to make sure you actually named your checkbox 'sig' in your form, and that it is spelled right, your form method = 'post' etc..

[RTS]

thanks, i figured it out. one more question though. I am trying to recieve the message ont eh other end, and if the column 'sig' says yes, I want the message to echo test on the bottom. this is what I have so far, but it is giving me a parse error.
[code]<?php

$con = mysql_connect("localhost","ZackBabtkis","") or die('Could not connect: ' . mysql_error());
mysql_select_db("test", $con);
$result = mysql_query("SELECT * FROM messages WHERE ID=" . $_GET['id']);

while($row = mysql_fetch_array($result))
{
$sigtest = $row['sig'];
$sig = if ($sigtest = "no") echo "test";
       
        echo $sig
?>[/code]
can someone tell me what might be wrong? I appreciate all the quick helpful replies by the way :)

[Psycho]

You need a semilcolon at the end of that last echo.

[.josh]

a) once again you have not properly used the ternary operator. The ternary operator format looks like this:

$variable = (condition) ? condition_is_true : condition_is_false;

b) you are also using an = instead of an == sign in your condition. Those are 2 different operators.  = is the assignment operator. == is the equality operator. 

c) you should NEVER insert a $_GET variable directly into a query, otherwise you are begging for sql injection hacks. you should always sanitize your variables before using them in queries. you should at the very very VERY least do something like what I have done below.

d) as mentioned above, you're missing a ; on your echo $sig line.

[code]
<?php

$con = mysql_connect("localhost","ZackBabtkis","") or die('Could not connect: ' . mysql_error());
mysql_select_db("test", $con);
$id = mysql_real_escape_string($_GET['id']);
$result = mysql_query("SELECT * FROM messages WHERE ID=" . $id);

while($row = mysql_fetch_array($result)) {
  $sigtest = $row['sig'];
  $sig = ($sigtest == "no") "test" : NULL; // will set $sig to "test" if $sigtest equals "no"
  echo $sig;
}
?>
[/code]