我们使用Java LDAP API通过LDAP将用户登录到Active Directory。我们希望增强登录功能，以进一步检查用户是否在给定的AD组中。有谁知道怎么做？
当前代码：
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 {
...
}
更新：请参阅以下解决方案。
答案 0 :(得分：18)
我们用下面的课程解决了这个问题。只需调用身份验证方法：
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;
}
}
答案 1 :(得分：14)
上述代码段中没有一个对我不起作用。在Google和tomcat源代码上花费1天后，以下代码可以很好地找到用户组。
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);
}
}
}
答案 2 :(得分：6)
最简单的方法是使用'lookup':(打开Ldap上下文：看上面的例子）
/**
* 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;
}
答案 3 :(得分：3)
查找用户是否是组成员的LDAP查找方法不正确，尤其是在您谈论登录用户时。对于实际登录组的用户，根据用户登录的计算机而不同。该列表需要包括来自域信任，嵌套组和本地组的组。
如果您正在寻找当前登录用户或您使用Java用户名和密码登录的用户的组成员身份，请尝试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() + ")");
}
答案 4 :(得分：2)
不确定Java API的具体细节，但执行此操作的一般方法是在查询/绑定中添加组检查。
答案 5 :(得分：1)
我无法使用java命名ldap为您提供工作代码。 我使用Spring LDAP，以及你的方式： 获取User对象，搜索sAMAccountName = USERNAME
等用户名获得对象后，您将检索属性memberOf - ＆gt;这将是一个列表，并检查Java中的特定一个。
这是我能想到的唯一方法。
答案 6 :(得分：1)
按照Sundaramurthi的回答，甚至可以以更简单的方式来完成此操作，即您无需查询所有用户组：
(&(objectClass=user)(sAMAccountName=XXXX)(memberOf=CN=YYY,OU=_Common-Access,OU=Groups,OU=_CORP,DC=XXX,DC=XX))
其中 XXXX-用户名 XXX.XX-域名 YYY-组名
这使您无论用户是否在组中都可以得到答案。
那就这样做吧
String userBase = "DC=XXX,DC=XX";
String CHECK_IF_USER_IN_GROUP = "(&(objectClass=user)(sAMAccountName=%s)(memberOf=CN=%s,OU=...,OU=...,OU=...,%s))";
String queryFilter = String.format(CHECK_IF_USER_IN_GROUP, user, group, userBase);
SearchControls constraints = new SearchControls();
constraints.setSearchScope(SearchControls.SUBTREE_SCOPE);
NamingEnumeration results = context.search(userBase, queryFilter, constraints);
if (results == null) {
throw new Exception("No answer from LDAP");
}
if (!results.hasMore()) {
System.out.println("No result found");
// user is not in the group
}
// user is in the group
答案 7 :(得分：1)
要在建立连接后查询用户是否属于给定组，您可以使用 Active Directory 中的 member 或 memberOf 属性进行查询。
使用 memberOf 属性： 使用的过滤器：(&(Group Member Attribute=Group DN)(objectClass=Group Object class)) 例如：(&(memberOf=CN=group,ou=qa_ou,dc=ppma,dc=org)(objectClass=group))
使用成员属性： 使用的过滤器 : (&(Group Member Attribute=User DN)(objectClass=Group Object class)) 例如：(&(member=CN=user,ou=qa_ou,dc=ppma,dc=org)(objectClass=group))
答案 8 :(得分：0)
不幸的是，答案因AD的安装以及其他类型的LDAP服务器而异。
我们必须解决同样的问题。最后，我们允许系统管理员向我们提供LDAP查询模式，我们将用户名（以及组名称，如果需要也是可变的）替换为模式。
答案 9 :(得分：0)
我觉得这很有用：
retrieves-group-membership for Active Directory
我有这段工作代码：
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.DirContext;
import javax.naming.directory.InitialDirContext;
import javax.naming.directory.SearchControls;
import javax.naming.directory.SearchResult;
import javax.naming.ldap.InitialLdapContext;
import javax.naming.ldap.LdapContext;
public class TestAD1 {
private static String userBase = "DC=SomeName,DC=SomeName,DC=SomeName,DC=SomeName,DC=COM,DC=US";
public static void main(String[] args) {
TestAD1 tad = new TestAD1();
try {
// Create a LDAP Context
Hashtable env = new Hashtable();
env.put(Context.INITIAL_CONTEXT_FACTORY, "com.sun.jndi.ldap.LdapCtxFactory");
env.put(Context.SECURITY_AUTHENTICATION, "simple");
env.put(Context.SECURITY_PRINCIPAL, "ldap.serviceaccount@domain.com");
env.put(Context.SECURITY_CREDENTIALS, "drowssap");
env.put(Context.PROVIDER_URL, "ldap://fully.qualified.server.name:389");
LdapContext ctx = new InitialLdapContext(env, null);
InitialDirContext inidircontext = new InitialDirContext(env);
DirContext dirctx = new InitialLdapContext(env, null);
System.out.println("Connection Successful.");
// Print all attributes of the name in namespace
SearchControls sctls = new SearchControls();
String retatts[] = {"sn", "mail", "displayName", "sAMAccountName"};
sctls.setReturningAttributes(retatts);
sctls.setSearchScope(SearchControls.SUBTREE_SCOPE);
String srchfilter = "(&(objectClass=user)(mail=*))";
String srchbase = userBase;
int totalresults = 0;
NamingEnumeration answer = dirctx.search(srchbase, srchfilter, sctls);
while (answer.hasMoreElements()) {
SearchResult sr = (SearchResult) answer.next();
totalresults++;
System.out.println(">>> " + sr.getName());
Attributes attrs = sr.getAttributes();
if (answer == null || !answer.hasMore()) {
System.out.println("No result found");
return;
}
if (attrs != null) {
try {
System.out.println(" surname: " + attrs.get("sn").get());
System.out.println(" Email - ID: " + attrs.get("mail").get());
System.out.println(" User - ID: " + attrs.get("displayName").get());
System.out.println(" Account Name: " + attrs.get("sAMAccountName").get());
tad.GetGroups(inidircontext, attrs.get("sAMAccountName").get().toString());
} catch (NullPointerException e) {
System.out.println("Error listing attributes..." + e);
}
}
System.out.println("Total Results : " + totalresults);
// close dir context
dirctx.close();
}
ctx.close();
} catch (NamingException e) {
System.out.println("Problem Search Active Directory..." + e);
//e.printStackTrace();
}
}
// Get all the groups.
public void GetGroups(InitialDirContext context, String username) throws NamingException {
String[] attrIdsToSearch = new String[]{"memberOf"};
String SEARCH_BY_SAM_ACCOUNT_NAME = "(sAMAccountName=%s)";
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;
}
SearchResult result = (SearchResult) results.next();
Attributes attrs = result.getAttributes();
Attribute attr = attrs.get(attrIdsToSearch[0]);
NamingEnumeration e = attr.getAll();
System.out.println(username + " is Member of the following groups : \n");
while (e.hasMore()) {
String value = (String) e.next();
System.out.println(value);
}
}
}
答案 10 :(得分：0)
此外，您可以通过以下方式修改接受的答案：Authenticating against Active Directory with Java on Linux：
String group="name of the group";
Iterator ig = groups.iterator();
Boolean bool=false;
while (ig.hasNext()) {
String a=ig.next().toString();
if (a.equals(group)) {
JOptionPane.showMessageDialog(this, "Authentication succeeded!");
bool=true;
// here you can do smth in case of success
}
}
if (bool==false){
JOptionPane.showMessageDialog(this, "Permission denied");
}