如何解决此Codeigniter-4登录控制器问题？

Question

我正在尝试在Codeigniter-4中创建一个基于电话和密码的登录，我已经创建了表单，这是我的控制器

  public function auth(){

    $model = new LoginModel();
    $user_phone = $this->request->getVar('user_phone');
    $password = $this->request->getVar('password');

    $result = $model->where("user_phone", "$user_phone");


if($result->numRows === 1){
    $user = $result->row_array();
    if($this->password->verify_hash($password, $user['password'])){
        unset($user['password']);

        return $user;

    } else {
       echo ('login');
    }
} else {
  echo ('login');
}

}

这是我的模型

class LoginModel extends Model
{
    protected $table = 'users';

    protected $allowedFields = ['username', 'password', 'user_phone'];

}

这是我的观点

        <?php
        $req_c       = $member["username"];


        ?>


          <div class="app-body__item-content padded">

              <div class="panel-body"><?= $req_c ?></div>

             ?>

          </div>

       <?php endforeach; ?>
       <?php endif; ?>

但它不会在视图中返回用户详细信息

Answer 1

我认为因为您在这里使用了双引号，所以它使您的$user_phone变量成为字符串

 $result = $model->where("user_phone", "$user_phone");

应该是这样的

 $result = $model->where("user_phone", $user_phone);

或者这个

 $result = $model->where(["user_phone"=> $user_phone]);