java 查询 AD 报错，请高手解决，急急急急

小弟最近在做一个系统的登录验证，要求从Windows 的 AD中验证，登录名为AD中的sAMAccountName的值，自己配了一台服务器，验证是没有问题的，但访问正式服务器时却抛出异常，请教各位高手，代码如下；import java.util.Hashtable;
import java.util.List;import javax.naming.AuthenticationException;
import javax.naming.Context;
import javax.naming.NamingEnumeration;
import javax.naming.NamingException;
import javax.naming.directory.DirContext;
import javax.naming.directory.InitialDirContext;
import javax.naming.directory.SearchControls;
import javax.naming.directory.SearchResult;
import javax.naming.ldap.Control;
import javax.naming.ldap.InitialLdapContext;/**
 * 验证用户登录
 * @throws NamingException 
 */
public String validateLogin() throws AMSException, NamingException {

String flag = "";
String responseMsg = "";
// 获取用户名，密码
String userName = this.getParameter("userName");

String password = this.getParameter("password");

String verifyCode = this.getParameter("verifyCode");// 文本框填写的验证码

Control[] connCtls = null;
  
    String userDN = "";
    
if(null==userName){
flag = "login";
responseMsg = "1";
}else if(null==password){
flag = "login";
responseMsg = "2";
}else if(null==verifyCode){
flag = "login";
responseMsg = "3";
}else{

ServletActionContext.getRequest().setAttribute("userName", userName);
verifyCode = verifyCode.toLowerCase();

AcitiveDirectoryConfig adConfig =new AcitiveDirectoryConfig();

//activedirectory ip
String activedirectoryIP=adConfig.getADConifg("AD_IP");

//activedirectory port
String activedirectoryPort=adConfig.getADConifg("AD_PORT");

String activedirectoryRoot=adConfig.getADConifg("AD_ROOT");

String verifyCodeConfirm=ServletActionContext.getRequest().getSession().getAttribute("random").toString();//servlet生成的验证码

String ad_username = adConfig.getADConifg("AD_USERNAME");

String ad_bindpassword = adConfig.getADConifg("AD_BINDPASSWORD");

Hashtable<String, String> env = new Hashtable<String, String>();

DirContext ctx;


env.put(Context.INITIAL_CONTEXT_FACTORY,"com.sun.jndi.ldap.LdapCtxFactory");
       
env.put(Context.PROVIDER_URL, "ldap://"+activedirectoryIP+":"+activedirectoryPort+"/"+activedirectoryRoot);//LDAP server
       
env.put(Context.SECURITY_AUTHENTICATION, "simple");

env.put(Context.SECURITY_PRINCIPAL, ad_username); 

env.put(Context.SECURITY_CREDENTIALS, ad_bindpassword);

        try{
               ctx = new InitialLdapContext(env,connCtls);
            SearchControls constraints = new SearchControls();
               constraints.setSearchScope(SearchControls.SUBTREE_SCOPE);
               System.out.println("*************************");
               NamingEnumeration en = ctx.search("", "sAMAccountName="+userName, constraints);---就这句报异常                
ctx.close();
               if(en == null){
                flag = "login";
                responseMsg = "4";
                System.out.println("11111111");
               }
               if(!en.hasMoreElements()){
                flag = "login";
               responseMsg = "4";
               System.out.println("222222222222");
               }else{
                while (en != null && en.hasMoreElements()){//maybe more than one element
                   Object obj = en.nextElement();
                   if(obj instanceof SearchResult){
                       SearchResult si = (SearchResult) obj;
                       userDN += si.getName();
                       userDN += "," + activedirectoryRoot;
                       System.out.println("userDN==========================="+userDN);
                       try {
               Hashtable<String, String> env1 = new Hashtable<String, String>();
              
             DirContext ctx1;
            
             env1.put(Context.PROVIDER_URL, "ldap://"+activedirectoryIP+":"+activedirectoryPort+"/"+activedirectoryRoot);
            
             env1.put(Context.SECURITY_AUTHENTICATION, "simple");
            
             env1.put(Context.SECURITY_PRINCIPAL, userDN); 
            
             env1.put(Context.SECURITY_CREDENTIALS, password);
            
             env1.put(Context.INITIAL_CONTEXT_FACTORY,"com.sun.jndi.ldap.LdapCtxFactory");
                          
             ctx1 = new InitialDirContext(env1);
            
             ctx1.close();
            
             ServletActionContext.getRequest().getSession().setAttribute("LoginUserName", userName);
            
             flag = "validateLogin";
            
                      }catch (AuthenticationException e) {
                     System.out.println("33333333333333333");
                       flag = "login";
                      responseMsg = "4";
                      }catch (NamingException e) {
                     System.out.println("44444444444444444");
                       flag = "login";
                      responseMsg = "5";
                      }
                   }
                   else{
                    System.out.println("5555555555555555");
                    flag = "login";
               responseMsg = "4";
                   }
               }
               }
              }catch(AuthenticationException e){
               System.out.println("666666666666666");
               flag = "login";
              responseMsg = "4";
              }catch (NamingException e) {
               logger.info("NamingException:"+e);
                   System.out.println("77777777777777777");----这里抛出
                     flag = "login";
                    responseMsg = "5";
                }
             
}
ServletActionContext.getRequest().setAttribute("responseMsg", responseMsg);
ServletActionContext.getRequest().getSession().setAttribute("computerinfo", userProcess.getCorp());
return flag;
}我的做法是先以管理员账号登录进去（好像不容许匿名登录），根据输入的用户名（即sAMAccountName）查到匹配的那条记录（如果有这条记录），得到该用户的cn后，再次连接AD，如果成功，则登录正确，请各位大大指教错误，异常如下
*************************
11-06-10 15:48:40 [INFO ] org.apache.log4j.Logger {UserAction.java:219} - Naming
Exception:javax.naming.NamingException: [LDAP: error code 1 - 000020D6: SvcErr:
DSID-031007DB, problem 5012 (DIR_ERROR), data 0
77777777777777777说明：测试服务器为2003   正式服务器为2008  （不知道有影响没）在线等结果，急