Cakephp 3 - 身份验证登录 Return 错误

Question

这段代码有什么问题？ Controller 和 Identify only return false！我的数据库列是 senha（密码）和电子邮件。而且我无法登录。我正在使用哈希，密码为 255 个字符，没问题。但它不起作用！

ContaController.php

public function initialize() {
    $this->loadComponent('Flash');
    $this->loadComponent('Auth', [
        'authenticate' => [
            'Form' => [
                'fields' => ['email' => 'email', 'senha' => 'senha'],
                'userModel' => 'Conta',
                'finder' => 'auth',
            ]
        ],
        'authorize' => ['Controller'],
        'loginAction' => [
            'controller' => 'Conta',
            'action' => 'index',
        ],
        'loginRedirect' => [
            'controller' => 'Conta',
            'action' => 'minha-agenda'
        ],
        'logoutRedirect' => [
            'controller' => 'Conta',
            'action' => 'index',
        ],
        'storage' => 'Memory'
    ]);
    $this->Auth->allow(['index']);
}


public function index() {
    if ($this->request->is('ajax') || $this->request->is('post')) {
        $user = $this->Auth->identify();
        if ($user) {
            $this->Auth->setUser($user);
            // return $this->redirect($this->Auth->redirectUrl());
            echo 'success';
        } else {
            var_dump($user);
            echo 'incorrect';

        }
    }
}

ContaTable.php

public function initialize(array $config) {
    parent::initialize($config);

    $this->setTable('alunos');
    $this->setDisplayField('id');
    $this->setPrimaryKey('id');

    $this->addBehavior('Timestamp');

    $this->belongsToMany('Alunos', [
        'foreignKey' => 'interesses_id',
        'targetForeignKey' => 'alunos_id',
        'joinTable' => 'alunos_interesses'
    ]);
}

public function validationDefault(Validator $validator) {
    $validator
        ->notEmpty('email', 'A username is required')
        ->notEmpty('senha', 'A password is required');
    return $validator;
}

public function findAuth(\Cake\ORM\Query $query, array $options) {
    $query
        ->select(['id', 'email', 'senha'])
        ->where(['Conta.email' => $options['email']])
        ->andWhere(['Conta.senha' => $options['senha']]);
    return $query;
}

我需要帮助解决这个问题。数据库中的列不同，所以我不打算将其用作默认值。再往前会变成ajax，但现在是这样，因为我无法解决！

Answer 1

将'fields' => ['email' => 'email', 'senha' => 'senha']更改为'fields' => ['username' => 'email', 'password' => 'senha']

使用此配置，您不需要自定义查找器，除非您需要 select 列。通过使用默认查找器，您可以随时取消设置属性。

确保您的登录表单控件和 table 列被命名为 email 和 senha

然后，我想知道你是否应该在 AppController 中加载 authcomponent。

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