can someone tell me what the error in this page? - php

Question

i get this error line on this php file . can someone locate where is the error ?
-------------You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1----------
i have this page for votes to users but if i vote in one user this vote goes to all users . how can i make this code when voting this vote goes only to its user .
--
// Connects to your Database
mysql_connect("localhost", "dbusername", "dbpassword") or die(mysql_error());
mysql_select_db("mydatabase") or die(mysql_error());
//We only run this code if the user has just clicked a voting link
if ( $mode=="vote")
{
//If the user has already voted on the particular thing, we do not allow them to vote again
//$cookie = "Mysite$id";
if(isset($_COOKIE[$cookie]))
{
Echo "Sorry You have already ranked that site <p>";
}
//Otherwise, we set a cooking telling us they have now voted
else
{
$month = 2592000 + time();
setcookie('Mysite'.$id, 'Voted', $month);
//Then we update the voting information by adding 1 to the total votes and adding their vote (1,2,3,etc) to the total rating
mysql_query ("UPDATE userads SET total = total+$voted, votes = votes+1 WHERE id = $id");
}
}
if ( $mode2=="vote")
{
//If the user has already voted on the particular thing, we do not allow them to vote again
//$cookie = "Mysite$id";
if(isset($_COOKIE[$cookie]))
{
Echo "Sorry You have already ranked that site <p>";
}
//Otherwise, we set a cooking telling us they have now voted
else
{
$month = 2592000 + time();
setcookie('Mysite'.$id, 'Voted', $month);
//Then we update the voting information by adding 1 to the total votes and adding their vote (1,2,3,etc) to the total rating
mysql_query ("UPDATE userads SET total = total+$voted, nvotes = nvotes+1 WHERE id = $id");
}
}
//Puts SQL Data into an array
$data = mysql_query("SELECT * FROM userads WHERE id = $id ") or die(mysql_error());
//Now we loop through all the data
while($ratings = mysql_fetch_array( $data ))
?>
<link href="style.css" type="text/css" rel="stylesheet" />
{
<?php
echo '<div id="voting_14" class="voting voting_template_votess-up-down">';
echo "<strong class='positive_votes'>";
$current = $ratings[votes];
echo "<span>+" . round($current,0) . "</span>";
echo " <input class='vote_positive' type='submit'>";
echo '</strong>';
echo "<strong class='negative_votes'>";
$current2 = $ratings[nvotes];
echo " <input class='vote_negative' type='submit'>";
echo "<span>-". round($current2,0) ."</span>";
echo '</strong>';
echo '</div>';
}
---the end
i have sql table userads with : id , name , username , total, votes , nvotes.

Answer 1

Correct your code to following,
setcookie('Mysite'.$id, 'Voted', $month); // ERROR 1
and
while($ratings = mysql_fetch_array( $data ))
{ // ERROR 2
?>

Answer 2

I copy pasted your code in a file and ran:
php -l your_script.php
Yields:
Parse error: syntax error, unexpected '}' in your_script.php on line 78
So, that last bracket } at the very end is causing a parse error. Either that or you didn't post the matching if/while/etc. in your post and the problem is elsewhere.

Answer 3

You have a syntax error. Probably your configuration doesn't display errors and you get a blank screen.

Answer 4

You are missing quotes here:
setcookie(Mysite.$id, Voted, $month);
It should be:
setcookie('Mysite'.$id, 'Voted', $month);

Answer 5

Apparently something is wrong with your query. You can check your SQL query with a simple:
$sql = "SELECT * FROM userads WHERE id = $id ";
echo $sql;
My first guess is, that $id is not set properly.