我想答案就在这里：http://forum.spring.io/forum/spring-projects/data/ldap/66894-objectsid-and-ldaptemplate
在第二个最后一个帖子中，他描述了你遇到的同样的问题。在上一篇文章中，他描述了一个修复方法，即将此添加到Bean配置文件中：

<bean id="contextSource" class="org.springframework.ldap.core.support.LdapContextSource">
    <property name="url" value="ldap://ldapserver.domain.com:389" />
    <property name="base" value="dc=domain,dc=com" />
    <property name="userDn" value="cn=binduser,cn=Users,dc=domain,dc=com" />
    <property name="password" value="bindpwd"/>
    <property name="baseEnvironmentProperties">
        <map>
        <entry key="java.naming.ldap.attributes.binary">
            <value>objectSid</value>
        </entry>
        </map>
    </property>
</bean>

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您必须修改域的值，但我认为重要的部分是baseEnvironmentProperties。
This thread还描述了一种编程方式来设置该属性（尽管对于objectGuid，但您可以只交换该属性）。

AbstractContextSource contextSource = (AbstractContextSource) ldapTemplate.getContextSource();
Map<String,String> baseEnvironmentProperties = new HashMap<String, String>();
baseEnvironmentProperties.put("java.naming.ldap.attributes.binary", "objectSid");
contextSource.setBaseEnvironmentProperties(baseEnvironmentProperties);
contextSource.afterPropertiesSet();

型

我通过在configure方法中添加环境属性来实现这一点：

@Configuration
@EnableWebSecurity
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {
@Override
protected void configure(HttpSecurity http) throws Exception {
    http
            .authorizeRequests()
            .antMatchers("/", "/home").permitAll()
            .anyRequest().authenticated()
            .and()
            .formLogin()
            .loginPage("/login")
            .permitAll()
            .and()
            .logout()
            .permitAll();
}
@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
    auth
            .authenticationProvider(activeDirectoryLdapAuthenticationProvider());
}
private ActiveDirectoryLdapAuthenticationProvider activeDirectoryLdapAuthenticationProvider() {
    ActiveDirectoryLdapAuthenticationProvider provider =
            new ActiveDirectoryLdapAuthenticationProvider(LdapConfig.AD_DOMAIN, LdapConfig.AD_SERVER);
// ************** NEW ENVIRONMENT PROPERTIES **********************************
    Map<String, Object> environmentProperties = new HashMap<>();
    environmentProperties.put("java.naming.ldap.attributes.binary", "objectsid");
    provider.setContextEnvironmentProperties(environmentProperties);
// ************** END OF NEW ENVIRONMENT PROPERTIES ***************************
    provider.setUserDetailsContextMapper(new LdapUserDetailsContextMapper());
    return provider;
    }
}

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然后在UserDetailContextMapper中像这样阅读它：

public class CustomUserDetailsContextMapper implements UserDetailsContextMapper {
private final static Logger logger = LoggerFactory.getLogger(CustomUserDetailsContextMapper.class);
@Override
public UserDetails mapUserFromContext(DirContextOperations ctx, String username, Collection<? extends GrantedAuthority> authorities) {
    logger.info(ctx.getDn().toString());
    byte[] byteSid = null;
    try {
        byteSid = (byte[]) ctx.getAttributes().get("objectsid").get();
    } catch (NamingException e) {
        e.printStackTrace();
    }
    String sid = LdapUtils.convertBinarySidToString(byteSid);
    logger.info("SID: {}", sid);
    return new User(username, "notUsed", true, true, true, true,
            AuthorityUtils.createAuthorityList("ROLE_USER"));    
}

型
我希望这是有帮助的！

这不是关于Spring的，而是关于普通的JavaLDAP访问（以防有人发现这个页面在寻找它--就像我一样）。在构建DirContext时，需要告诉它将objectSid保持为二进制形式。否则，它将尝试将字节解码为String，这将用unicode REPLACEMENT CHARACTER U+FFFD（�）替换无效字符-从而无法获得正确的Sid。

DirContext ldapConnect() throws NamingException {
    Hashtable<String, String> env = new Hashtable<>();
    env.put(Context.INITIAL_CONTEXT_FACTORY, "com.sun.jndi.ldap.LdapCtxFactory");
    env.put(Context.PROVIDER_URL, ldapServer);
    env.put(Context.SECURITY_AUTHENTICATION, "simple");
    env.put(Context.SECURITY_PRINCIPAL, ldapUser);
    env.put(Context.SECURITY_CREDENTIALS, ldapPassword);
    env.put("java.naming.ldap.attributes.binary", "objectSid"); // !!!!!!!!!!!!!!!!!!
    return new InitialDirContext(env);
}

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这基本上是Gabriel Lucis的答案，只是为了香草Java而不是Spring。