java 查询 AD 报错，请高手解决，该如何解决

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

[解决办法]
我也报这个错 不知道解决没