mysqli_query期望参数1为mysqli

时间:2015-08-21 09:07:10

标签: php mysql mysqli parameters

我有以下代码:

<?
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>";

?>

它仍然给我这个错误:

  

警告:mysqli_query()要求参数1为mysqli,第48行/home/u705139137/public_html/header.php中给出的字符串

     

警告:mysqli_fetch_array()要求参数1为mysqli_result,在第49行的/home/u705139137/public_html/header.php中给出null

     

警告:mysqli_query()期望参数1为mysqli,第51行/home/u705139137/public_html/header.php中给出的字符串

     

警告:mysqli_fetch_array()要求参数1为mysqli_result,在第53行的/home/u705139137/public_html/header.php中给出null

     

警告:mysqli_query()要求参数1为mysqli,第57行/home/u705139137/public_html/header.php中给出的字符串

     

警告:mysqli_fetch_array()要求参数1为mysqli_result,在第58行的/home/u705139137/public_html/header.php中给出null

我做错了什么?

2 个答案:

答案 0 :(得分:0)

更改你的mysqli_query,如下所示

 <?
    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>";

    ?>

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

答案 1 :(得分:0)

这样做

$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