Question

我在从MySQL切换到MySQLi时遇到问题。我使用MySQL尝试了以下代码，它工作正常。当我切换到MySQLi时，它返回一个空值。

MySQL（工作正常）

public function Login_User($username, $password) 
{
    $md5_password = md5($password);
    $query = mysql_query("SELECT uid FROM users WHERE username='$username' and password='$md5_password'");
    if(mysql_num_rows($query) == 1)
    {
        $row=mysql_fetch_array($query);
        return $row['uid'];
    }
}

MySQLi（$ numrows为空）

public function Login_User($username, $password) 
{
    $md5_password = md5($password);
    $query = mysqli_query($connection, "SELECT uid FROM users WHERE username='$username' and password='$md5_password'");
    $numrows = mysqli_num_rows($query);
    echo "username: " . $username . ", Password: " . $password . "<br/>";
    echo "rows: " . $numrows;
    if($numrows == 1)
    {
        $row = mysqli_fetch_array($query, MYSQLI_ASSOC);
        return $row['uid'];
    }
}

数据库连接（MySQL）

$dbc = @mysql_connect ('localhost','root','')
   or die ('Could not connect to MySQL: ' . mysql_error() );

mysql_select_db ('socialhouse')
  or die ('problems with selecting database: ' . mysql_error() );

数据库连接（MySQLi）

$connection = mysqli_connect('localhost', 'root', '', 'socialhouse');

if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();}

有人知道我的MySQLi代码有什么问题吗？提前致谢

Answer 1

因为您必须使用MySQLi扩展将$connection变量传递到脚本中，所以您现在正尝试访问此函数范围之外的变量。将它作为参数传递给你的函数，你应该好好去。

Answer 2

这是基本的想法：

        $mysqli = new mysqli(.......);
        $stmt = $mysqli->prepare("SELECT ..."); 
        $stmt->bind_param('isi..', ...);
        $stmt->execute();   
        $stmt->store_result(); 
        $numrows=$stmt->num_rows;
        print "query returned " . $numrows;

MySQL到MySQLi（获取num_rows）

2 个答案: