dongqu1783 2013-01-28 18:56 采纳率: 100%
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致命错误:在[重复]中调用非对象的成员函数query()

Possible Duplicate:
Call to a member function query() on a non-object in query()?

I am getting the

Fatal error: Call to a member function query() on a non-object Error when trying to get row count from an Account table in my database. Here is the code:

$link = mysqli_connect("localhost", "Username", "Password", "Database");    
        if ($result = $mysqli->query($link, "SELECT * FROM Accounts WHERE Username=" . $_POST['EmailTbx'] . " AND Password=" . $_POST['PasswordTbx'] . "")){
            $field_cnt = $result->field_count;
            echo $field_cnt;
            $result->close();
        }
        $mysqli->close();
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2条回答 默认 最新

  • douzhan8303 2013-01-28 19:02
    关注

    It looks like you got your variables mixed up. Try the following (does not prevent injection):

    $link = mysqli_connect("localhost", "Username", "Password", "Database");    
            if ($result = $link->query("SELECT * FROM Accounts WHERE Username='{$_POST['EmailTbx']}' AND Password='{$_POST['PasswordTbx']}'")){
                $field_cnt = $result->field_count;
                echo $field_cnt;
                $result->close();
            }
            $mysqli->close();
    

    Try the following with escaping:

    $link = mysqli_connect("localhost", "Username", "Password", "Database");
    $email = mysqli_real_escape_string($link, $_POST['EmailTbx']);
    $pass = mysqli_real_escape_string($link, $_POST['PasswordTbx']);
            if ($result = $link->query("SELECT * FROM Accounts WHERE Username='{$email}' AND Password='{$pass}'")){
                $field_cnt = $result->field_count;
                echo $field_cnt;
                $result->close();
            }
            $mysqli->close();
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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