error when wanted to validate key on database - php

Question

i had the problem when i want to check my valid key on my database and showing this error
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\ravaaa\indexx.php on line 20
the db is login, tables is users (5) id,name,key,login
this the code
$error='';
$key="123";
$con = mysqli_connect("localhost","root","","login");
if (mysqli_connect_error())
{
echo "Failed to connect to MySQL: " . mysql_connect_error();
}
echo "sukses\n\n";
$query = mysqli_query($con, "SELECT * FROM users WHERE key='$key'");
$rows = mysqli_num_rows($query);
if($rows == 1){
echo "key is valid";
}
else
{
$error = "key is Invalid";
echo $error;
}
mysqli_close($con);

Answer 1

Are you sure your table have an column named "key"? Key is a reserved MySQL Keyword, and when i tried to create a table with this column to test i got an error, and after change the name "key" to another name (on the table and on the .php) i got no errors, just the message "key is Invalid".
Try to change the column name "key" and change on the php too, and check if you still get errors.

Answer 2

The column name key is a reserved word and therefore must be escaped using backticks:
select * from users WHERE `key` = '$key'
Without the escape, the query will fail and return FALSE per the mysqli_query docs:
Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or
EXPLAIN queries mysqli_query() will return a mysqli_result object. For
other successful queries mysqli_query() will return TRUE.
You can exit early with an error using the or die pattern:
mysqli_query($con, "SELECT * FROM users WHERE `key` = '$key'")
or die(mysqli_error($con));