Question

为什么findone（）将所有公共属性返回为null。

class User extends \yii\db\ActiveRecord 
{
    public $id;
    public $username;
    public $password;
    public $authKey;
    public $accessToken;


    /**
     * Finds user by username
     *
     * @param  string      $username
     * @return static|null
     */
    public static function findByUsername($username)
    {
        $result =  static::findOne(['username' => $username]);
        var_dump($result);
        return $result;
    }

返回

object(app\models\User)[81]
  public 'id' => null
  public 'username' => null
  public 'password' => null
  public 'authKey' => null
  public 'accessToken' => null
  private '_attributes' (yii\db\BaseActiveRecord) => 
    array (size=6)
      'id' => int 1
      'username' => string 'admin' (length=5)
      'password' => string '123456' (length=6)
      'auth_key' => string 'jkkk' (length=4)
      'created' => null
      'modified' => null

Answer 1

您只需从模型中删除db属性：

class User extends \yii\db\ActiveRecord 
{

    public static function findByUsername($username)
    {
        ....

Yii会自动在Active Record中为关联表的每一列定义一个属性。 您不应重新声明任何属性。

了解详情：http://www.yiiframework.com/doc-2.0/guide-db-active-record.html

yii2 find返回null

1 个答案: