Question

所以，我有一个名为''members'的集合，我的每个'成员'都有一个用户名和密码。我需要知道的是如何检查两者是否匹配，即用户名+密码=成功

这是我试过的，它正确搜索，如果没有用户，它就不会返回错误

public function userlogin($data)
    {
        $this->data = $data;
        $collection = $this->db->members;
        // find everything in the collection
        $cursor = $collection->find(array("username"=>$this->data['username'], "password"=>$this->data['password']));

        $test = array();
            // iterate through the results
            while( $cursor->hasNext() ) {
                $test[] = ($cursor->getNext());
            }
        //Print Results 

        if($test == NULL)
        {
            print "Sorry we are not able to find you";
            die;
        }
        //print json_encode($test);

    }

Answer 1

假设用户名/密码组合是唯一的，您可以使用 findOne ：

$mongoConn = new Mongo();
$database = $mongoConn->selectDB('myDatabase');
$collection = $database->selectCollection('members');
$user = $collection->findOne(array('username' => $username,'password' => $password));

如果要限制返回某些字段的数据，可以在findOne的末尾指定它们：

$user = $collection->findOne(array('username' => $username,'password' => $password),array('_id','firstname','lastname'));

Answer 2

这样的事情：

$mongo = new Mongo();
$db = $mongo->dbname;    

$user = $db->collection->findOne(array("username" => $username, "password" => $password));
if ($user->count() > 0)
    return $user;
return null;

或者：

$user = $db->collection->findOne(array("username" => $username, "password" => $password));
$user->limit(1);
if ($user->count(true) > 0)
    return $user;
return null;

简单的PHP mongoDB用户名和密码检查站点

2 个答案: