Java LDAP: ¿determina si el usuario está en un grupo determinado?

active-directory

Question

Iniciamos sesión en Active Directory a través de LDAP utilizando la API Java LDAP. Queremos mejorar nuestra funcionalidad de inicio de sesión para verificar si el usuario está en un grupo de AD determinado. ¿Alguien sabe como hacer esto?

Código actual:

import javax.naming.*;
import javax.naming.ldap.*;
LdapContext ctx = null;
Hashtable env = new Hashtable();
env.put(Context.INITIAL_CONTEXT_FACTORY,"com.sun.jndi.ldap.LdapCtxFactory");
env.put(Context.SECURITY_AUTHENTICATION,"simple");
env.put(Context.PROVIDER_URL, Config.get("ldap-url"));
try {
Control[] connCtls = new Control[] {new FastBindConnectionControl()};
ctx = new InitialLdapContext(env, connCtls);
ctx.addToEnvironment(Context.SECURITY_PRINCIPAL, "DOMAIN//" + username);
ctx.addToEnvironment(Context.SECURITY_CREDENTIALS, password);
ctx.reconnect(connCtls);
/* TODO: Only return true if user is in group "ABC" */
return true; //User authenticated
} catch (Exception e) {
return false; //User could NOT be authenticated
} finally {
...
}

Actualización : vea la solución a continuación.

Answer 1

Desafortunadamente, la respuesta varía según las instalaciones de AD y otros tipos de servidor LDAP.

Tuvimos que resolver el mismo problema. Al final permitimos que el administrador del sistema nos proporcione un patrón de consulta LDAP donde sustituimos el patrón con el nombre de usuario (y el nombre del grupo, si eso también debe ser variable).

Answer 2

No estoy seguro sobre los aspectos específicos de la API de Java, pero la forma genérica de hacerlo es agregar una verificación de grupo a la consulta / enlace.

Answer 3

No puedo darle un código de trabajo usando java name ldap. Utilicé Spring LDAP y la manera en que lo haces: Obtienes el objeto User, haces una búsqueda en el nombre de usuario algo así como sAMAccountName = USERNAME

Después de obtener el objeto, recupera la propiedad memberOf -> esto será una lista y buscará una específica en Java.

Esa es la única forma en que se me ocurre.

Answer 4

Solucionamos esto con la clase de abajo. Simplemente llame al método de autenticación:

import java.text.MessageFormat;
import java.util.*;
import javax.naming.*;
import org.apache.log4j.Level;
public class LdapGroupAuthenticator {
public static final String DISTINGUISHED_NAME = "distinguishedName";
public static final String CN = "cn";
public static final String MEMBER = "member";
public static final String MEMBER_OF = "memberOf";
public static final String SEARCH_BY_SAM_ACCOUNT_NAME = "(SAMAccountName={0})";
public static final String SEARCH_GROUP_BY_GROUP_CN = "(&(objectCategory=group)(cn={0}))";
/*
* Prepares and returns CN that can be used for AD query
* e.g. Converts "CN=**Dev - Test Group" to "**Dev - Test Group"
* Converts CN=**Dev - Test Group,OU=Distribution Lists,DC=DOMAIN,DC=com to "**Dev - Test Group"
*/
public static String getCN(String cnName) {
if (cnName != null && cnName.toUpperCase().startsWith("CN=")) {
cnName = cnName.substring(3);
}
int position = cnName.indexOf('','');
if (position == -1) {
return cnName;
} else {
return cnName.substring(0, position);
}
}
public static boolean isSame(String target, String candidate) {
if (target != null && target.equalsIgnoreCase(candidate)) {
return true;
}
return false;
}
public static boolean authenticate(String domain, String username, String password) {
Hashtable env = new Hashtable();
env.put(Context.INITIAL_CONTEXT_FACTORY,"com.sun.jndi.ldap.LdapCtxFactory");
env.put(Context.PROVIDER_URL, "ldap://1.2.3.4:389");
env.put(Context.SECURITY_AUTHENTICATION, "simple");
env.put(Context.SECURITY_PRINCIPAL, domain + "//" + username);
env.put(Context.SECURITY_CREDENTIALS, password);
DirContext ctx = null;
String defaultSearchBase = "DC=DOMAIN,DC=com";
String groupDistinguishedName = "DN=CN=DLS-APP-MyAdmin-C,OU=DLS File Permissions,DC=DOMAIN,DC=com";
try {
ctx = new InitialDirContext(env);
// userName is SAMAccountName
SearchResult sr = executeSearchSingleResult(ctx, SearchControls.SUBTREE_SCOPE, defaultSearchBase,
MessageFormat.format( SEARCH_BY_SAM_ACCOUNT_NAME, new Object[] {username}),
new String[] {DISTINGUISHED_NAME, CN, MEMBER_OF}
);
String groupCN = getCN(groupDistinguishedName);
HashMap processedUserGroups = new HashMap();
HashMap unProcessedUserGroups = new HashMap();
// Look for and process memberOf
Attribute memberOf = sr.getAttributes().get(MEMBER_OF);
if (memberOf != null) {
for ( Enumeration e1 = memberOf.getAll() ; e1.hasMoreElements() ; ) {
String unprocessedGroupDN = e1.nextElement().toString();
String unprocessedGroupCN = getCN(unprocessedGroupDN);
// Quick check for direct membership
if (isSame (groupCN, unprocessedGroupCN) && isSame (groupDistinguishedName, unprocessedGroupDN)) {
Log.info(username + " is authorized.");
return true;
} else {
unProcessedUserGroups.put(unprocessedGroupDN, unprocessedGroupCN);
}
}
if (userMemberOf(ctx, defaultSearchBase, processedUserGroups, unProcessedUserGroups, groupCN, groupDistinguishedName)) {
Log.info(username + " is authorized.");
return true;
}
}
Log.info(username + " is NOT authorized.");
return false;
} catch (AuthenticationException e) {
Log.info(username + " is NOT authenticated");
return false;
} catch (NamingException e) {
throw new SystemException(e);
} finally {
if (ctx != null) {
try {
ctx.close();
} catch (NamingException e) {
throw new SystemException(e);
}
}
}
}
public static boolean userMemberOf(DirContext ctx, String searchBase, HashMap processedUserGroups, HashMap unProcessedUserGroups, String groupCN, String groupDistinguishedName) throws NamingException {
HashMap newUnProcessedGroups = new HashMap();
for (Iterator entry = unProcessedUserGroups.keySet().iterator(); entry.hasNext();) {
String unprocessedGroupDistinguishedName = (String) entry.next();
String unprocessedGroupCN = (String)unProcessedUserGroups.get(unprocessedGroupDistinguishedName);
if ( processedUserGroups.get(unprocessedGroupDistinguishedName) != null) {
Log.info("Found : " + unprocessedGroupDistinguishedName +" in processedGroups. skipping further processing of it..." );
// We already traversed this.
continue;
}
if (isSame (groupCN, unprocessedGroupCN) && isSame (groupDistinguishedName, unprocessedGroupDistinguishedName)) {
Log.info("Found Match DistinguishedName : " + unprocessedGroupDistinguishedName +", CN : " + unprocessedGroupCN );
return true;
}
}
for (Iterator entry = unProcessedUserGroups.keySet().iterator(); entry.hasNext();) {
String unprocessedGroupDistinguishedName = (String) entry.next();
String unprocessedGroupCN = (String)unProcessedUserGroups.get(unprocessedGroupDistinguishedName);
processedUserGroups.put(unprocessedGroupDistinguishedName, unprocessedGroupCN);
// Fetch Groups in unprocessedGroupCN and put them in newUnProcessedGroups
NamingEnumeration ns = executeSearch(ctx, SearchControls.SUBTREE_SCOPE, searchBase,
MessageFormat.format( SEARCH_GROUP_BY_GROUP_CN, new Object[] {unprocessedGroupCN}),
new String[] {CN, DISTINGUISHED_NAME, MEMBER_OF});
// Loop through the search results
while (ns.hasMoreElements()) {
SearchResult sr = (SearchResult) ns.next();
// Make sure we''re looking at correct distinguishedName, because we''re querying by CN
String userDistinguishedName = sr.getAttributes().get(DISTINGUISHED_NAME).get().toString();
if (!isSame(unprocessedGroupDistinguishedName, userDistinguishedName)) {
Log.info("Processing CN : " + unprocessedGroupCN + ", DN : " + unprocessedGroupDistinguishedName +", Got DN : " + userDistinguishedName +", Ignoring...");
continue;
}
Log.info("Processing for memberOf CN : " + unprocessedGroupCN + ", DN : " + unprocessedGroupDistinguishedName);
// Look for and process memberOf
Attribute memberOf = sr.getAttributes().get(MEMBER_OF);
if (memberOf != null) {
for ( Enumeration e1 = memberOf.getAll() ; e1.hasMoreElements() ; ) {
String unprocessedChildGroupDN = e1.nextElement().toString();
String unprocessedChildGroupCN = getCN(unprocessedChildGroupDN);
Log.info("Adding to List of un-processed groups : " + unprocessedChildGroupDN +", CN : " + unprocessedChildGroupCN);
newUnProcessedGroups.put(unprocessedChildGroupDN, unprocessedChildGroupCN);
}
}
}
}
if (newUnProcessedGroups.size() == 0) {
Log.info("newUnProcessedGroups.size() is 0. returning false...");
return false;
}
// process unProcessedUserGroups
return userMemberOf(ctx, searchBase, processedUserGroups, newUnProcessedGroups, groupCN, groupDistinguishedName);
}
private static NamingEnumeration executeSearch(DirContext ctx, int searchScope, String searchBase, String searchFilter, String[] attributes) throws NamingException {
// Create the search controls
SearchControls searchCtls = new SearchControls();
// Specify the attributes to return
if (attributes != null) {
searchCtls.setReturningAttributes(attributes);
}
// Specify the search scope
searchCtls.setSearchScope(searchScope);
// Search for objects using the filter
NamingEnumeration result = ctx.search(searchBase, searchFilter,searchCtls);
return result;
}
private static SearchResult executeSearchSingleResult(DirContext ctx, int searchScope, String searchBase, String searchFilter, String[] attributes) throws NamingException {
NamingEnumeration result = executeSearch(ctx, searchScope, searchBase, searchFilter, attributes);
SearchResult sr = null;
// Loop through the search results
while (result.hasMoreElements()) {
sr = (SearchResult) result.next();
break;
}
return sr;
}
}

Answer 5

Los métodos de búsqueda de LDAP para encontrar si un usuario es miembro de un grupo no son correctos, especialmente si está hablando de un usuario que ha iniciado sesión. Para un usuario que está realmente conectado, la lista de grupos varía según la computadora en la que el usuario inició sesión. Esa lista debe incluir grupos de trusts de dominio, grupos anidados y grupos locales.

Si está buscando membresías grupales del usuario que está actualmente conectado o un usuario que está iniciando sesión con un nombre de usuario y contraseña en Java, pruebe Waffle .

IWindowsAuthProvider prov = new WindowsAuthProviderImpl();
IWindowsIdentity identity = prov.logonUser("username", "password");
System.out.println("User identity: " + identity.getFqn());
for(IWindowsAccount group : identity.getGroups()) {
System.out.println(" " + group.getFqn() + " (" + group.getSidString() + ")");
} 

Answer 6

Ninguno de los fragmentos de código anteriores no funcionó para mí. Después de 1 día, gastar en Google y Tomcat source following code funcionó bien para encontrar grupos de usuarios.

import java.util.Hashtable;
import javax.naming.CompositeName;
import javax.naming.Context;
import javax.naming.Name;
import javax.naming.NameParser;
import javax.naming.NamingEnumeration;
import javax.naming.NamingException;
import javax.naming.directory.Attribute;
import javax.naming.directory.Attributes;
import javax.naming.directory.InitialDirContext;
import javax.naming.directory.SearchControls;
import javax.naming.directory.SearchResult;
public class MemberOfTest{
private static final String contextFactory = "com.sun.jndi.ldap.LdapCtxFactory";
private static final String connectionURL = "ldap://HOST:PORT";
private static final String connectionName = "CN=Query,CN=Users,DC=XXX,DC=XX";
private static final String connectionPassword = "XXX";
// Optioanl
private static final String authentication = null;
private static final String protocol = null;
private static String username = "XXXX";
private static final String MEMBER_OF = "memberOf";
private static final String[] attrIdsToSearch = new String[] { MEMBER_OF };
public static final String SEARCH_BY_SAM_ACCOUNT_NAME = "(sAMAccountName=%s)";
public static final String SEARCH_GROUP_BY_GROUP_CN = "(&(objectCategory=group)(cn={0}))";
private static String userBase = "DC=XXX,DC=XXX";
public static void main(String[] args) throws NamingException {
Hashtable<String, String> env = new Hashtable<String, String>();
// Configure our directory context environment.
env.put(Context.INITIAL_CONTEXT_FACTORY, contextFactory);
env.put(Context.PROVIDER_URL, connectionURL);
env.put(Context.SECURITY_PRINCIPAL, connectionName);
env.put(Context.SECURITY_CREDENTIALS, connectionPassword);
if (authentication != null)
env.put(Context.SECURITY_AUTHENTICATION, authentication);
if (protocol != null)
env.put(Context.SECURITY_PROTOCOL, protocol);
InitialDirContext context = new InitialDirContext(env);
String filter = String.format(SEARCH_BY_SAM_ACCOUNT_NAME, username);
SearchControls constraints = new SearchControls();
constraints.setSearchScope(SearchControls.SUBTREE_SCOPE);
constraints.setReturningAttributes(attrIdsToSearch);
NamingEnumeration results = context.search(userBase, filter,constraints);
// Fail if no entries found
if (results == null || !results.hasMore()) {
System.out.println("No result found");
return;
}
// Get result for the first entry found
SearchResult result = (SearchResult) results.next();
// Get the entry''s distinguished name
NameParser parser = context.getNameParser("");
Name contextName = parser.parse(context.getNameInNamespace());
Name baseName = parser.parse(userBase);
Name entryName = parser.parse(new CompositeName(result.getName())
.get(0));
// Get the entry''s attributes
Attributes attrs = result.getAttributes();
Attribute attr = attrs.get(attrIdsToSearch[0]);
NamingEnumeration e = attr.getAll();
System.out.println("Member of");
while (e.hasMore()) {
String value = (String) e.next();
System.out.println(value);
}
}
}

Answer 7

La forma más fácil es con ''búsqueda'': (para abrir un contexto Ldap: mira los ejemplos anteriores)

 /**
* Tests if an Active Directory user exists in an Active Directory group.
* @param ctx LDAP Context.
* @param dnADGroup distinguishedName of group.
* @param dnADUser distinguishedName of user.
* @return True if user is member of group.
*/
public static boolean isMemberOfADGroup(LdapContext ctx, String dnADGroup, String dnADUser) {
try {
DirContext lookedContext = (DirContext) (ctx.lookup(dnADGroup));
Attribute attrs = lookedContext.getAttributes("").get("member");
for (int i = 0; i < attrs.size(); i++) {
String foundMember = (String) attrs.get(i);
if(foundMember.equals(dnADUser)) {
return true;
}
}
} catch (NamingException ex) {
String msg = "There has been an error trying to determin a group membership for AD user with distinguishedName: "+dnADUser;
System.out.println(msg);
ex.printStackTrace();
}
return false;
}