我在我的应用程序中使用FOSUserBundle。我想通过HTTP服务做两件事:
检查密码。该服务可能如下所示(密码不会被加密):
public function checkPasswordValidity($userId, $password) {
$user = $this->getDoctrine()
->getRepository('MyCompany\UserBundle\Entity\User')
->find($userId);
if (specialFunction($user , $password))
echo 'Valid Password';
else
echo 'Invalid Password';
}
通过其他HTTP服务创建新用户。参数将是用户名和密码。
答案 0 :(得分:62)
检查密码:
$encoder_service = $this->get('security.encoder_factory');
$encoder = $encoder_service->getEncoder($user);
$encoded_pass = $encoder->encodePassword($password, $user->getSalt());
//then compare $user->getPassword() and $encoded_pass
创建新用户:
$userManager = $this->get('fos_user.user_manager');
$user = $userManager->createUser();
$user->setUsername($login);
$user->setPlainPassword($pass);
...
$userManager->updateUser($user);
答案 1 :(得分:6)
对我而言:
$encoder_service = $this->get('security.encoder_factory');
$encoder = $encoder_service->getEncoder($user);
if ($encoder->isPasswordValid($user->getPassword(), $password, $user->getSalt()) {}
我没有测试第二个问题,但我认为它已经是answered了。
答案 2 :(得分:0)
登录 phpmyadmin ,访问 fos_user_user 表,点击插入>填写字段,用户名,电子邮件,角色等。
使用此php脚本生成salt和密码:
<?php
$salt = base_convert(sha1(uniqid(mt_rand(), true)), 16, 36);
echo "Salt used: " . $salt ."<br/>";
echo "<br/>";
$password = 'adminpasswordhere';
$salted = $password.'{'.$salt.'}';
$digest = hash('sha512', $salted, true);
for ($i=1; $i<5000; $i++) {
$digest = hash('sha512', $digest.$salted, true);
}
$encodedPassword = base64_encode($digest);
echo "Password used: " . $password ."<br/>";
echo "<br/>";
echo "Encrypted Password: " . $encodedPassword ."<br/>";
?>