Question

我希望通过从以下结构化的LDAP存储库查询其ID来获取用户属性列表

dn: uid=E000001 ,ou=People,o=Company,o=Internal
cn: BOB DOLE
statusid: active
memberof: cn=foo_group, cn=Foos, ou=Groups, o=Company,o=Internal
memberof: cn=bar_group, cn=Foos, ou=Groups, o=Company,o=Internal

dn: uid=E000002 ,ou=People,o=Company,o=Internal
cn: MARK TEST
statusid: active
memberof: cn=foo_group, cn=Foos, ou=Groups, o=Company,o=Internal
memberof: cn=bar_group, cn=Foos, ou=Groups, o=Company,o=Internal

例如，我查询用户ID＆＃34; E00001＆＃34;。我想回复这个

["cn=foo_group, cn=Foos, ou=Groups, o=Company,o=Internal", "cn=bar_group, cn=Foos, ou=Groups, o=Company,o=Internal"

Answer 1

以下是检索用户组的多种方法：

如果你有一个没有嵌套组的简单LDAP服务器，memberOf通常就足够了：

String userCN = "user1";

//Get the attribute of user's "memberOf"
ArrayList<?> membersOf = ldapTemplate.search(
        query().where("sAMAccountName").is(userCN),
        (AttributesMapper<ArrayList<?>>) attrs -> Collections.list(attrs.get("memberOf").getAll())
).get(0);

但是如果你有嵌套组，事情会变得更复杂：

/*
 * Get user distinguised name, example: "user" -> "CN=User Name,OU=Groups,OU=Domain Users,DC=company,DC=something,DC=org"
 * This will be used for our query later
 */
String distinguishedName = ldapTemplate.search(
        query().where("sAMAccountName").is(userCN),
        (AttributesMapper<String>) attrs -> attrs.get("distinguishedName").get().toString()
).get(0); //.get(0): we assume that search will return a result 

/*
 * This one recursively search for all (nested) group that this user belongs to
 * "member:1.2.840.113556.1.4.1941:" is a magic attribute, Reference: 
 * https://msdn.microsoft.com/en-us/library/aa746475(v=vs.85).aspx
 * However, this filter is usually slow in case your ad directory is large.
 */
List<String> allGroups = ldapTemplate.search(
        query().searchScope(SearchScope.SUBTREE)
                .where("member:1.2.840.113556.1.4.1941:").is(distinguishedName),
        (AttributesMapper<String>) attrs -> attrs.get("cn").get().toString()
);

对于googlers：请注意，它具有org.springframework.boot:spring-boot-starter-data-ldap的依赖关系，以防有人需要Bean初始化代码：

@Component
@EnableConfigurationProperties
public class Ldap {
    @Bean
    @ConfigurationProperties(prefix="ldap.contextSource")
    public LdapContextSource contextSource() {
        return new LdapContextSource();
    }

    @Bean
    public LdapTemplate ldapTemplate(ContextSource contextSource) {
        return new LdapTemplate(contextSource);
    }
}

在application.yml中使用以下配置模板：

ldap:
    contextSource:
        url: ldap://your-ldap.server
        base: dc=Company,dc=Domain,dc=Controller
        userDn: username
        password: hunter2
        #you'll want connection polling set to true so ldapTemplate reuse the connection when searching recursively
        pooled: true

当魔术数字的表现不好时：如果你的ldap目录很大，那么使用幻数的最后一个实际上很慢，在这种情况下，递归搜索ldap会更快。这是一个帮助类，可以详尽地搜索用户所属的所有组：

public class LdapSearchRecursive {
    private final LdapTemplate ldapTemplate;
    private Set<String> groups;

    public LdapSearchRecursive(LdapTemplate ldapTemplate) {
        this.ldapTemplate = ldapTemplate;
        this.groups = new HashSet<>();
    }

    /**
     * Retrieve all groups that this user belongs to.
     */
    public Set<String> getAllGroupsForUserRecursively(String userCN) {
        List<String> distinguishedNames = this.ldapTemplate.search(
                query().where("objectCategory").is("user").and(
                        query().where("sAMAccountName").is(userCN)
                                .or(query().where("userPrincipalName").is(userCN))
                ),
                (AttributesMapper<String>) attrs -> attrs.get("distinguishedName").get().toString()
        );

        if (distinguishedNames.isEmpty()) {
            throw new UsernameNotFoundException("User not recognized in LDAP");
        }

        return this.getAllGroupsRecursivelyByUserDistinguishedName(distinguishedNames.get(0), null);
    }

    private Set<String> getAllGroupsRecursivelyByUserDistinguishedName(String dn, @Nullable String parentDN) {
        List<String> results = this.ldapTemplate.search(
                query().where("member").is(dn),
                (AttributesMapper<String>) attrs -> attrs.get("distinguishedName").get().toString()
        );

        for (String result : results) {
            if (!(result.equals(parentDN) //circular, ignore
                    || this.groups.contains(result) //duplicate, ignore
                    )) {
                this.getAllGroupsRecursivelyByUserDistinguishedName(result, dn);
            }
        }
        this.groups.addAll(results);

        return this.groups;
    }
}

Spring Boot / Spring LDAP获取用户成员列表

1 个答案: