错误：警告：mysqli_fetch_array（）要求参数1为mysqli_result，第15行/var/www/html/data.php中给出布尔值[复制]

Question

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

I am trying to select data from a MySQL table to show data's in my c# datagridview, but I get one of the following error messages:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /var/www/html/data.php on line 15

This is my PHP code:

<?php 

$servername = "********";
$username = "root";
$password = "pws";
$dbname = "dbs";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$sql = "SELECT userid,password,first_name,last_name,role,active FROM aster_users";
$result = mysql_query($sql);
while($row = mysqli_fetch_array($result)) 
    {
        $userid=$row['userid'];
        $password=$row['password'];
        $first_name=$row['first_name'];
        $last_name=$row['last_name'];
        $role=$row['role'];
        $active=$row['active'];
    }
$conn->close();
?>

Help me,

</div>

Answer 1

You cannot mix the mysql and mysqli. For database queries please use mysqli_query($conn,$query). http://php.net/manual/en/mysqli.query.php

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$sql = "SELECT userid,password,first_name,last_name,role,active FROM aster_users";
$result = mysqli_query($conn,$sql); // NOTICE HOW THIS IS MYSQLI QUERY. NOT MYSQL
while($row = mysqli_fetch_array($result)) 
    {
        $userid=$row['userid'];
        $password=$row['password'];
        $first_name=$row['first_name'];
        $last_name=$row['last_name'];
        $role=$row['role'];
        $active=$row['active'];
    }
mysqli_close($conn); //Use Procedural Code Here. not $conn->close();