简单的PHP mongoDB用户名和密码检查网站

Question

所以，我有一个名为'会员'的集合，并且我有我的每个'成员'的用户名和密码。 我需要知道的是我怎么检查，看看两者是否匹配，即用户名+密码=成功

这是我曾尝试并正确进行搜索，只是它没有返回错误，如果没有用户

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

事情是这样的：

$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; 

Answer 2

假设一个用户名/密码组合是唯一的，你可以使用一个findOne：

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

如果你想限制回来某些字段中的数据，你可以在findOne结束指定它们：

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