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2009年3月6日 星期五

[Java] 建立由 LDAP AD 認證的機制

From: Polin Wei

環境:
OS : Windows Server 2k3
Domain: mydomain.com
Host : x.x.x.x(此為指定 ip address)

以 Java 程式撰寫 對 LDAP-AD 的認證, 程式碼如下:

import javax.naming.Context;
import javax.naming.InitialContext;
import javax.naming.NamingEnumeration;
import javax.naming.NamingException;
import javax.naming.directory.Attribute;
import javax.naming.directory.Attributes;
import javax.naming.directory.BasicAttribute;
import javax.naming.directory.DirContext;
import javax.naming.directory.InitialDirContext;
import javax.naming.directory.ModificationItem;
import javax.naming.directory.SearchControls;
import javax.naming.directory.SearchResult;
import javax.naming.ldap.InitialLdapContext;
import javax.naming.ldap.LdapContext;
 /**
     * AD LDAP 登入認證
     *
     * @param ldap_url   like ldap://x.x.x.x:389/DC=mydomain,DC=com
     * @param account
     * @param password
     * @return String[0] array 0 :0 success,1 fail,2 LDAP connect fail,3 unknow
     */
    public String[] LDAP_AUTH_AD(String ldap_url, String account, String password) {
        String[] returnStr = new String[2];
        Hashtable<String, String> env = new Hashtable<String, String>();
        env.put(Context.INITIAL_CONTEXT_FACTORY, "com.sun.jndi.ldap.LdapCtxFactory");
        env.put(Context.PROVIDER_URL, ldap_url);
        env.put(Context.SECURITY_AUTHENTICATION, "simple");
        env.put(Context.SECURITY_PRINCIPAL, account+"@mydomain.com");
        env.put(Context.SECURITY_CREDENTIALS, password);

        LdapContext ctx = null;
        try {
            ctx = new InitialLdapContext(env, null);
            returnStr[0] = "0";
        } catch (javax.naming.AuthenticationException e) {
            returnStr[0] = "1";

            return returnStr;
        } catch (javax.naming.CommunicationException e) {
            // System.out.println("Can't connect to ldap server!");
            returnStr[0] = "2";

            return returnStr;
        } catch (Exception e) {
            System.out.println("error");
            e.printStackTrace();
            returnStr[0] = "3";

            return returnStr;
        } finally {
            if (ctx != null) {
                try {
                    ctx.close();
                } catch (NamingException e) {

                }
            }
        }
    }

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