Use an IF STATEMENT with INSERT IGNORE - php

Question

I have a voting system, which users can use to vote on posts. Thing is, they can't vote twice on the same post. Which is good. I did that like this:
$query = "INSERT IGNORE INTO vote_id (vote, postid, userid)
VALUES ('1', '{$postid}', '{$userid}')";
$result = mysqli_query($db, $query);
header('Location: index.php');
(The $db is my variable that connects to the database.)
They can still click on the vote, even if it doesn't insert into the table. The problem is, when it is IGNORE they shouldn't be able to click.
How do I separate the INSERT and IGNORE, so I can use the IGNORE in an if statement? Or in any other way, that can change the layout of the page?

Answer 1

Without code showing "vote" it's hard to say, but if it's button, then you can simply query a SELECT for a user_id and post_id, and if it finds at least one row, then add disabled to button. Like that:
<button name="vote" value="1" disabled>Vote</button>
Or just don't display anything "to vote" at all when query returns rows.
Other way is to check session, but it'd let the same user to vote again after relogging.

Answer 2

Thing is, they can't vote twice on the same post. Which is good. I did that like this:
What you've shown us here doesn't prevent people from voting twice (a unique index on postid and userid will - but that's not in your post). I do hope you are cleaning those parameters before splicing them into a an SQL statement like that.
But that is a comment.
How do I separate the INSERT and IGNORE, so I can use the IGNORE in an if statement
While I'm not 100% certain, I would expect mysqli_affected_rows() to return 0 if no rows were inserted. A more robust solution would be to omit the 'IGNORE' from the SQL statement and handle the duplicate key error:
$result = mysqli_query($db, $query);
$error=mysqli_errno($db);
switch ($error) {
case 0:
// all good
header('Location: index.php');
break;
case 1062:
print "Please don't vote twice\n";
exit;
default:
trigger_error("Query failed " . $query . " // " . mysqli_error($db));
print "whoops";
exit;
}