访问通过解码JSON字符串创建的对象属性时的空值

Question

In my PHP script I need to decode a Json string and then transfer the decoded value to a class. Something like:

index.php

$params = json_decode('input');
$obj = new User();
$obj->setParams($params);
$obj->Register();

class.php

class User{
private $mParams;

public function setParams($params)
    $mParams = $params;
}

public function Register(){
    $username = $mParams->{'username'};
    $password = $mParams->{'password'};
}
....
}

The problem is, in Register(), when I print the $username and $password, I just got NULL. But I'm sure the $params decoded from Json is not NULL because, if I print it in setParams, I can get username and password. And, if I directly transfer the $params to Register() everything is fine.

So I feel strange that why I can not set the $params to the class's member and then call the class's member function to access it.

Thanks,

Answer 1

$this->mParams->username
 ^^^^^^

You want to access the object's property, so use $this . Also, no need for the string-in-brackets syntax.

Answer 2

When setting the member variables of an object in php, you must prefix $this-> to distinguish them from local variables. You want:

class User{
  private $mParams;

  public function setParams($params)
      $this->mParams = $params;
  }

  public function Register(){
      $username = $this->mParams->{'username'};
      $password = $this->mParams->{'password'};
      echo $username . " : " . $password;
  }
  ....
}