Php 如何在LoginForm中更改表用户

我在询问Yii2中的LoginForm

安装了Yii之后，我得到了默认的Web，其中包含登录表单。此表单将连接到表名“user”

然后我修改默认设置，以创建具有不同登录表单的新网站。我还为登录名“db_user”创建了新表。我仍然使用commons/model中名为“LoginForm”的默认模型进行登录。这是密码

<?php
namespace common\models;

use Yii;
use yii\base\Model;

/**
 * Login form
 */
class LoginForm extends Model
{
public $username;
public $password;
public $rememberMe = true;

private $_user = false;


/**
 * @inheritdoc
 */
public function rules()
{
    return [
        // username and password are both required
        [['username', 'password'], 'required'],
        // rememberMe must be a boolean value
        ['rememberMe', 'boolean'],
        // password is validated by validatePassword()
        ['password', 'validatePassword'],
    ];
}

/**
 * Validates the password.
 * This method serves as the inline validation for password.
 *
 * @param string $attribute the attribute currently being validated
 * @param array $params the additional name-value pairs given in the rule
 */
public function validatePassword($attribute, $params)
{
    if (!$this->hasErrors()) {
        $user = $this->getUser();
        if (!$user || !$user->validatePassword($this->password)) {
            $this->addError($attribute, 'Incorrect username or password.');
        }
    }
}

/**
 * Logs in a user using the provided username and password.
 *
 * @return boolean whether the user is logged in successfully
 */
public function login()
{
    if ($this->validate()) {
        return Yii::$app->user->login($this->getUser(), $this->rememberMe ? 3600 * 24 * 30 : 0);
    } else {
        return false;
    }
}

/**
 * Finds user by [[username]]
 *
 * @return User|null
 */
public function getUser()
{
    if ($this->_user === false) {
        $this->_user = User::findByUsername($this->username);
    }

    return $this->_user;
}


 }

---

LoginForm模型只是表单的模型，而不是用户模型

User::findByUsername($this->username);

如您所见，LoginForm中使用的db模型是User，如果您进入文件models/User.php，您将看到一行：

public static function tableName()
{
    return '{{%user}}';
}

改为：

public static function tableName()
{
    return 'db_user';
}

---

祝你好运

试试这个怎么做——我不明白为什么我会在投票中落选。至少先解释一下