mysqli_fetch_assoc（）正好需要1个参数，给定2个

Question

Only 1 parameter is needed in fetch assoc but I'm checking to see if the username is found and it gives me an error because it only expects one parameter.

function user_exists($username) {
   $connection = mysqli_connect("localhost","root","","c3392262");
   $username = sanitize($username);
   $query = mysqli_query($connection,"SELECT COUNT(`user_id`) FROM 'users' WHERE `username` =           '$username'");
 return (mysqli_fetch_assoc($query, 0) == 1) ? true : false;
}

Answer 1

Like the error says, mysqli_fetch_assoc() only expects 1 parameter, but you're passing 2. Plus the function returns either an associative array or NULL, so you should rather test its return value against NULL:

return (mysqli_fetch_assoc($query)!=NULL)?true:false

On a side note, you don't really need to use brackets, since != has a higher priority than ?: already

return mysqli_fetch_assoc($query)!=NULL?true:false

Plus, as pointed out by @rar, you don't really need to use the ternary operator either, since the comparison between mysqli_fetch_assoc() and NULL "returns" a boolean value already

return mysqli_fetch_assoc($query)!=NULL