我正在使用 Spring3.1 和 hibernate4 开发项目。
现在我想从属性文件中加密敏感数据,如用户名、数据库密码。
这是我遵循的步骤:(参考http://www.jasypt.org/spring31.html)
1.配置占位符:
<bean id="propertyConfigurer" class="org.jasypt.spring31.properties.EncryptablePropertyPlaceholderConfigurer">
<constructor-arg ref="configurationEncryptor" />
<property name="locations">
<list>
<value>/WEB-INF/classes/connection.properties</value>
</list>
</property>
<property name="ignoreUnresolvablePlaceholders" value="true"/>
</bean>
2.加密器的配置
<bean id="encryptorConfig" class="org.jasypt.encryption.pbe.config.SimplePBEConfig">
<property name="algorithm" value="PBEWithMD5AndDES" />
<property name="password" value="MASTERPASSWORD" />
</bean>
<bean id="configurationEncryptor" class="org.jasypt.encryption.pbe.StandardPBEStringEncryptor">
<property name="config" ref="encryptorConfig" />
</bean>
3.数据库连接
<bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
<property name="driverClassName" value="${data.driver}"/>
<property name="url" value="${data.url}"/>
<property name="username" value="${data.user}"/>
<property name="password" value="${data.password}"/>
</bean>
4.通过命令使用jasypt生成加密值:
encrypt input="MY_DATABASE_PASSWORD" password="MASTERPASSWORD" algorithm="PBEWithMD5ANDDES"
5.connection.properties 文件包含
data.user=ENC(VO0A3aXAu71CCgzGFa+nJO/7M/0b5MF2)
data.password=ENC(EogzgPllaXTDm7wq5kRp6uPmkWq6pmDV)
当我运行应用程序时,我仍然收到错误消息:
org.postgresql.util.PSQLException: FATAL: password authentication failed for user "ENC(VO0A3aXAu71CCgzGFa+nJO/7M/0b5MF2)"
这些是我为将 Spring 应用程序与 jasypt 集成而包含的额外 jar:
commons-dbcp-1.1.jar
commons-lang-2.1.jar
commons-pool-1.2.jar
icu4j-3.4.4.jar
jasypt-1.9.0.jar
jasypt-1.9.0-lite.jar
jasypt-acegisecurity-1.9.0.jar
jasypt-hibernate4-1.9.0.jar
jasypt-spring3-1.9.0.jar
jasypt-spring31-1.9.0.jar
我错过了什么还是有任何 jar 兼容性问题?