尝试连接数据库时出现mysqli_query错误[重复]

I am getting this error:

Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\xampp\htdocs\php_programs\Query_Print_Results.php on line 18

It fails on this statement:

$result = mysqli_query($con,$query) or die(mysql_error());

I checked the value of $con and it is Resource ID = 6. Why? How do I get around this? Never had this happen before, but this is the first time I am not developing on localhost.

<?PHP
include 'Login_Parameters.php';

$query = "SELECT * FROM t_589";

$con = mysql_connect($DB_HOST,$DB_USER, $DB_PASS, $DB_DATABASE);

//* Check the Connection                                                *

  if (mysqli_connect_errno())
  {
    echo "Connect failed: " . mysqli_connect_error();
    exit();
  }
  echo "Connected to $DB_DATABASE<br />";


$result = mysqli_query($con,$query) or die(mysql_error());

/* Get field information for all columns */
$finfo = mysqli_fetch_fields($result);

foreach ($finfo as $val) {
  echo "<th>" . $val->name . "</th>";
}

echo "</tr>";
while($row = mysqli_fetch_array($result)){
  echo "<tr>";
  for($i = 0; $i < mysqli_num_fields($result); $i++){
    echo "<td>". $row[$i] ."</td>";
  }
  echo "</tr>";
}

echo "</table>";

 // Free result set
 mysqli_free_result($result);
 ?>

</div>

Perhaps you mean to call mysqli_connect() instead?

$con = mysqli_connect($DB_HOST,$DB_USER, $DB_PASS, $DB_DATABASE);

And as an aside, later on in your code, you need to make one other change to use mysqli_error().

$result = mysqli_query($con,$query) or die(mysqli_error());

You have to change this line:

$con = mysql_connect($DB_HOST,$DB_USER, $DB_PASS, $DB_DATABASE);

To this:

$con = mysqli_connect($DB_HOST,$DB_USER, $DB_PASS, $DB_DATABASE);