mysqli_query ожидает параметр 1, чтобы быть MySQLi

Question

У меня есть следующий код:

<? 
include"mysql.php"; 
$getusers= mysqli_query ("select count(*) as cnt from users",$c); 
$totalusers= mysqli_fetch_array ($getusers); 
$totalusers=$totalusers['cnt']; 
$getusers= mysqli_query ("select * from users",$c); 
$totalmoney=0; 
while($usr= mysqli_fetch_array ($getusers)) 
{ 
$totalmoney=$totalmoney+$usr['total_earned']; 
} 
$getoffers= mysqli_query ("select count(*) as cnt from offers where   active=1",$c); 
$ocnt= mysqli_fetch_array ($getoffers); 
$ocnt=$ocnt['cnt']; 

print"<p>Total Offers: $ocnt<br>Total Members: $totalusers<br>Total Earned: \$$totalmoney</p>"; 

?> 

еще это дает мне эту ошибку:

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in /home/u705139137/public_html/header.php on line 48

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/u705139137/public_html/header.php on line 49

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in /home/u705139137/public_html/header.php on line 51

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/u705139137/public_html/header.php on line 53

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in /home/u705139137/public_html/header.php on line 57

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/u705139137/public_html/header.php on line 58

Что я сделал не так?

Answer 1

Изменение mysqli_query как ниже

<? 
    include"mysql.php"; 
    $getusers= mysqli_query ($c,"select count(*) as cnt from users"); 
    $totalusers= mysqli_fetch_array ($getusers); 
    $totalusers=$totalusers['cnt']; 
    $getusers= mysqli_query ("select * from users",$c); 
    $totalmoney=0; 
    while($usr= mysqli_fetch_array ($getusers)) 
    { 
    $totalmoney=$totalmoney+$usr['total_earned']; 
    } 
    $getoffers= mysqli_query ("select count(*) as cnt from offers where   active=1",$c); 
    $ocnt= mysqli_fetch_array ($getoffers); 
    $ocnt=$ocnt['cnt']; 

    print"<p>Total Offers: $ocnt<br>Total Members: $totalusers<br>Total Earned: \$$totalmoney</p>"; 

    ?> 

Look; http://www.w3schools.com/php/func_mysqli_query.asp

Answer 2

ли это как то

$mysqli = new mysqli("localhost", "my_user", "my_password", "world"); 
$getusers= mysqli_query ($mysqli, "select count(*) as cnt from users"); 

http://php.net/manual/en/mysqli.query.php