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I got this below warning while executing the code. How to overcome that warning error. Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\login®ister\core\function\users.php on line 6.
<?php
function user_exists($username)
{
$username = sanitize($username);
$query=mysqli_query("SELECT COUNT (`user_id`) FROM `users` WHERE `username`='$username' ");
return (mysqli_result($query, 0)==1) ? true : false ;
}
function user_active($username)
{
$username = sanitize($username);
$query=mysqli_query("SELECT COUNT (`user_id`) FROM `users` WHERE `username`='$username' AND `active`=1 ");
return (mysql_result($query, 0) ==1 ) ? true : false ;
}
?>
</div>
The procedural style of mysqli_query requires a link identifier as first parameter.
mysqli_query($link, 'query here');
Don't forget you're inside a function there as well, any $link set outside the function is unknown inside the function.