php - 使用模型中的函数

Question

用户.php（模型）：

...
public function hashPassword($password){
    $hashed = hash('sha256', $password . self::HASH_CODE);
    return $hashed;
}
...

用户身份.php

...
else if($user->password!==Users::hashPassword($this->password))
...

错误：

Non-static method Users::hashPassword() should not be called statically, assuming $this from incompatible context

score 1 · Accepted Answer

您需要将其定义hashPassword()为静态函数才能调用它Users::hashPassword()：

public static function hashPassword($password) {
    ...

否则，您可以创建Users该类的实例并以非静态方式调用它：

$users = new Users();
$users->hashPassword($password);

严格yii来说，您可以使用以下方式调用它（取决于您的设置）：

Yii::app()->Users->hashPassword($password);

score 1 · Accepted Answer

1

else if($user->password!==Users::model()->hashPassword($this->password))

这不是静态方法

于 2012-11-01T03:36:11.407 回答

score 1 · Accepted Answer

制作函数static

public static function hashPassword($password){
    $hashed = hash('sha256', $password . self::HASH_CODE);
    return $hashed; }

3 回答 3