Symfony 3登录阵列的电话 - Symfony 3 login by array of phones

Symfony 3 login by array of phones

本文关键字：电话阵列登录 Symfony | 更新日期: 2023-09-27

大家好!我想登录我的用户，这是一个微不足道的任务。但问题是检查用户数组的值!

例如，我有实体用户和电话。用户有多部手机。我需要登录所有手机的用户。如何使用安全包的默认工具?

我没有发现任何像我这样的问题，并且阅读了所有关于Symfony安全的文档。我唯一想做的是创建自定义提供程序。但我不认为这能解决我的问题。

亲爱的symfoners，有什么想法吗?:)

你必须在security中设置一个安全提供者。yml

security:
    # http://symfony.com/doc/current/book/security.html#where-do-users-come-from-user-providers
...
    providers:
        my_db_provider:
            entity:
                class: AppBundle:User
...

那么你的实体AppBundle:User应该实现接口Symfony'Component'Security'Core'User'UserInterface，并有一个自定义存储库，例如UserRepository实现接口Symfony'Bridge'Doctrine'Security'User'UserLoaderInterface。你的UserRepository类应该是这样的:

<?php
use Doctrine'ORM'NonUniqueResultException;
use Doctrine'ORM'NoResultException;
use AppBundle'Entity'User;
use Symfony'Bridge'Doctrine'Security'User'UserLoaderInterface;
use Symfony'Component'Security'Core'Exception'UsernameNotFoundException;
class UserRepository extends EntityRepository implements UserLoaderInterface
{
 /**
  * Loads the user for the given username.
  *
  * This method must return null if the user is not found.
  *
  * @param string $username The username
  * @return null|Utilisateur
  * @throws 'Exception
  */
  public function loadUserByUsername($username)
  {
    //Here you write a custom query to retrive the user base on the fields you require. 
    // Here I have used username, email and phone number
    $q = $this
        ->createQueryBuilder('u')
        ->select('u')
        ->leftJoin('u.phones', 'p')
        ->where('u.username = :username or u.email= :email or p.phoneNumber= :phone')
        ->setParameter('username', $username)
        ->setParameter('email', $username)
        ->setParameter('phone ', $username)
        ->getQuery();
    try {
        $user = $q->getSingleResult();
    } catch (NoResultException $e) {
        throw new UsernameNotFoundException(sprintf('Unable to find an active user AppBundle:User object identified by "%s".', $username), 0, $e);
    } catch (NonUniqueResultException $ex) {
        throw new 'Exception("The user you provided is not unique");
    }
    return $user;
  }
}

你的AppByndle:User实体类应该是这样的:

<?php

  use Doctrine'Common'Collections'ArrayCollection;
  use Doctrine'Common'Collections'Collection;
  use Doctrine'ORM'Mapping as ORM;
  use Symfony'Component'Security'Core'User'UserInterface;
  /**
   * User
   *
   * @ORM'Table(name="user")
   * @ORM'Entity(repositoryClass="AppBundle'Dao'UserRepository")
 */
 class User implements UserInterface
 {
  /**
   * @var integer
   *
   * @ORM'Column(name="id", type="integer", nullable=false)
   * @ORM'Id
   * @ORM'GeneratedValue(strategy="IDENTITY")
   */
   private $id;
   /**
    * @var string
    *
    * @ORM'Column(name="username", type="string", length=254, nullable=false, unique=true)
    */
    private $username;
    ....
    ....
  }