php - 如何解决我的编译器警告

Question

完整代码:

<?php 

if (isset($_POST['usname']) && isset($_POST['pasw'])){

    $username = mysql_ramznegar($_POST['usname'],$link);
    $password = mysql_ramznegar($_POST['pasw'],$link);

$query = "SELECT * FROM `admin` WHERE `usname` = $username";
    $admin_result = mysql_query($query,$link);
    $admin_result = mysql_fetch_array($query,MYSQL_ASSOC);

    if($admin_result['usname']==$username && $admin_result['passw']==$password){


            if($admin_result[usname] == "admin"){

            $_SESSION['auth'] = 1;

                redirect("panel");

                   exit;                                

    }
     else{

        echo 'ur username or password is invalid';
        redirect("index.php");
     }


}

}

?>

我有一个数据库，其中的表(admin)有两列，第一列是usname，第二列是passw，当我输入时，我仍然选择查询用户名和密码编译器向我显示此错误，如何解决此编译器警告:

Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in C:\xampp\htdocs\ramznegar1\index.php on line 60

You should change

$admin_result = mysql_fetch_array($query, MYSQL_ASSOC)

To

$admin_result = mysql_fetch_array($admin_result, MYSQL_ASSOC)

我编辑了代码，但再次遇到此错误:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\ramznegar1\index.php on line 60

新代码是:http://pastebin.com/ycNT0XqM

Answer 1

你应该改变:

$query = "SELECT * FROM `admin` WHERE `usname` = $username";
$admin_result = mysql_query($query,$link);
$admin_result = mysql_fetch_array($query,MYSQL_ASSOC);

至

$query = "SELECT * FROM `admin` WHERE `usname` = $username"; 
$admin_result = mysqli_query($query,$link); 
$admin_result = mysqli_fetch_array($admin_result,MYSQL_ASSOC);

参见http://php.net/manual/en/function.mysql-fetch-array.php有关 mysql_fetch_array 如何工作的信息