Java 无法加密配置文件中的密码

Cannot encrypt password in configuration file

提问by nachokk

I'm having trouble encrypting the database password in hibernate.cfg.xml

我在加密数据库密码时遇到问题 hibernate.cfg.xml

This is my property file.

这是我的财产文件。

<!-- Database connection settings -->
<property name="connection.driver_class">com.microsoft.sqlserver.jdbc.SQLServerDriver</property>
<property name="connection.url">jdbc:sqlserver://localhost:1433;databaseName=TEST;</property>
<property name="connection.username">sa</property>
<!-- Encryption -->
<property name="connection.password">ENC(vMO/j5jfpaU2cUhPVoOk5Q==)</property>
<property name="connection.provider_class">org.jasypt.hibernate4.connectionprovider.EncryptedPasswordDriverManagerConnectionProvider</property>
<property name="connection.encryptor_registered_name">hibernateEncryptor</property>

Then in the HiberanteUtil.javaI have this

然后在HiberanteUtil.java我有这个

// Builds session factory.
private static SessionFactory configureSessionFactory() 
    throws HibernateException {

  Configuration configuration = new Configuration().configure();
  StandardPBEStringEncryptor encryptor =
      new StandardPBEStringEncryptor();
  encryptor.setPassword("pass");

  HibernatePBEEncryptorRegistry registry =
      HibernatePBEEncryptorRegistry.getInstance();

  registry.registerPBEStringEncryptor("hibernateEncryptor", encryptor);

  ServiceRegistry serviceRegistry = new ServiceRegistryBuilder()
      .applySettings(configuration.getProperties()).buildServiceRegistry();

  return configuration.buildSessionFactory(serviceRegistry);
}

I've created the encrypted password with encrypt.bat.

我已经用encrypt.bat.

Then the error i have is

然后我的错误是

com.microsoft.sqlserver.jdbc.SQLServerException: Login failed for user 'sa'. ClientConnectionId:8033573f-5f52-4fe9-a728-fbe4f57d89c4

com.microsoft.sqlserver.jdbc.SQLServerException：用户“sa”登录失败。ClientConnectionId:8033573f-5f52-4fe9-a728-fbe4f57d89c4

If I remove this part

如果我删除这部分

StandardPBEStringEncryptor encryptor =
        new StandardPBEStringEncryptor();
encryptor.setPassword("someKey");
HibernatePBEEncryptorRegistry registry =
        HibernatePBEEncryptorRegistry.getInstance();

registry.registerPBEStringEncryptor(
        "hibernateEncryptor", encryptor);

I have the same error, so I think it doesn't register but I have no idea how to do it.

我有同样的错误，所以我认为它没有注册，但我不知道该怎么做。

This is how i encrypt

这就是我加密的方式

UPDATE

更新

The only thing i can made to get it work is something like this, but is not the way i think.

我唯一能让它工作的事情就是这样，但不是我想的那样。

StandardPBEStringEncryptor encryptor =
                new StandardPBEStringEncryptor();
        encryptor.setPassword("somePass");
        encryptor.setAlgorithm("PBEWITHMD5ANDDES");
        String pass=encryptor.decrypt("HhpmA/XmJoLro8TYYu4YyA==");
        HibernatePBEEncryptorRegistry registry =
                HibernatePBEEncryptorRegistry.getInstance();
        registry.registerPBEStringEncryptor(
                "hibernateEncryptor", encryptor);

        Configuration configuration = new Configuration().configure()
                .setProperty("hibernate.connection.encryptor_registered_name","hibernateEncryptor")
                .setProperty("hibernate.connection.password",pass);

So i think the problem is with the "hibernateEncryptor", i think i need to register

所以我认为问题出在"hibernateEncryptor"，我想我需要注册

  <typedef name="encryptedString" class="org.jasypt.hibernate4.type.EncryptedStringType">
   <param name="encryptorRegisteredName">hibernateEncryptor</param>
  <typedef>

But when i put it in hibernate.cfg.xmlsays invalid mapping, so i add it to a class with annotation but nothing happen cause i think this is read after database connection that is what i want to encrypt. :(

但是当我把它放在hibernate.cfg.xml说无效映射中时，所以我将它添加到一个带有注释的类中，但没有发生任何事情，因为我认为这是在数据库连接后读取的，这是我想要加密的。:(

@TypeDef(name="encryptedString",typeClass=org.jasypt.hibernate4.type.EncryptedStringType.class,
        parameters= {@Parameter(name="encryptorRegisteredName",value="hibernateEncryptor")})

采纳答案by nachokk

This is not the proper way to do it but solves.

这不是正确的方法，而是解决了。

StandardPBEStringEncryptor encryptor =new StandardPBEStringEncryptor();
encryptor.setPassword("somePass");
encryptor.setAlgorithm("PBEWITHMD5ANDDES");
Configuration configuration = new Configuration().configure();
String pass=encryptor.decrypt(configuration.getProperty("hibernate.connection.password"));
configuration.setProperty("hibernate.connection.password",pass);   

And in hibernate.cfg

而在 hibernate.cfg

    <property name="connection.username">sa</property>
    <property name="connection.password">Nzuyhu5PJJwsVH3mdw==</property>

回答by Tim Spann

http://www.jasypt.org/hibernate.html

http://www.jasypt.org/hibernate.html

Why not switch algorithms to: PBEWithMD5AndTripleDES

为什么不将算法切换到：PBEWithMD5AndTripleDES

Take a look at this post on StackOverflow: Error implementing Jasypt with Hibernate 3 and Struts 2

看看 StackOverflow 上的这篇文章： 使用 Hibernate 3 和 Struts 2 实现 Jasypt 时出错

回答by B Black

You might try this:

你可以试试这个：

StandardPBEStringEncryptor strongEncryptor = new StandardPBEStringEncryptor();
strongEncryptor.setPassword("jasypt");
strongEncryptor.setAlgorithm("PBEWITHMD5ANDDES");
HibernatePBEEncryptorRegistry registry =                          HibernatePBEEncryptorRegistry.getInstance();
registry.registerPBEStringEncryptor("strongHibernateStringEncryptor", strongEncryptor);

Configuration configuration = new Configuration();
configuration.configure("hibernate.cfg.xml");
configuration.setProperty("hibernate.connection.password", strongEncryptor.decrypt(configuration.getProperty("hibernate.connection.password")));
ServiceRegistryBuilder serviceRegistryBuilder = new ServiceRegistryBuilder().applySettings(configuration.getProperties());
sessionFactory = configuration.buildSessionFactory(serviceRegistryBuilder.buildServiceRegistry());