LDAP:错误代码 49 - 简单绑定失败:NT_STATUS_LOGON_FAILURE

LDAP: error code 49 - Simple Bind Failed: NT_STATUS_LOGON_FAILURE(LDAP:错误代码 49 - 简单绑定失败:NT_STATUS_LOGON_FAILURE)

问题描述

我正在尝试对用户进行身份验证，但它抛出 Exception.可能是配置有问题.

I am trying to authenticate the user but it throws Exception.May be there is problem in configuration.

public class LdapApplication {
private static final String INITIAL_CONTEXT_FACTORY = "com.sun.jndi.ldap.LdapCtxFactory";
private static final String SECURITY_AUTHENTICATION ="simple";
private static final String NAMED_CONTEXT = "CN=Users";
private static final String SAM_ACCOUNT_NAME = "sAMAccountName=";

public static void main(String[] args) {

    Hashtable env = new Hashtable();

    env.put(Context.INITIAL_CONTEXT_FACTORY,INITIAL_CONTEXT_FACTORY);
    env.put(Context.PROVIDER_URL, "ldap://ip:portNo/dc=organisation,dc=in");
    env.put(Context.SECURITY_AUTHENTICATION, SECURITY_AUTHENTICATION);
    env.put(Context.SECURITY_PRINCIPAL, "cn=userName,cn=Users");
    env.put(Context.SECURITY_CREDENTIALS, "password" );
    DirContext context = null;

    NamingEnumeration namingEnumeration = null;
    try {
        context = new InitialDirContext(env);

        namingEnumeration = context.search(NAMED_CONTEXT, SAM_ACCOUNT_NAME+ userName, null);
        while (namingEnumeration.hasMore()) {
            SearchResult searchResult = (SearchResult) namingEnumeration.next();
            Attributes attributes = searchResult.getAttributes();

            System.out.println(" Person Common Name = " + attributes.get("cn"));
          System.out.println(" Person Display Name = " + attributes.get("displayName"));

            }catch(Exception e){
                System.out.println(e.getMessage());
                e.printStackTrace();

            }
        }
    } catch (Throwable e) {
        e.printStackTrace();
    } finally {
        if (namingEnumeration != null) {
            try {
                namingEnumeration.close();
            } catch (Exception e) {
            }
        }
        if (context != null) {
            try {
                context.close();
            } catch (Exception e) {
            }
        }
    }

}

}

但如果我提到 Context.SECURITY_PRINCIPAL 作为 "organisation\userName" 而不是 "cn=userName,cn=Users" 它工作得很好.请提出一个可能的解决方案，因为我的要求是使用 cn 或 dc 给 SECURITY_PRINCIPAL 一些东西.

but if i mention Context.SECURITY_PRINCIPAL as "organisation\userName" instead of "cn=userName,cn=Users" it works perfectly fine. Kindly suggest a possible solution because my requirement is to give SECURITY_PRINCIPAL something using cn or dc.

推荐答案

我们在代码中遇到了同样的问题，我们通过在用户名前添加域名来修复它.不要输入 user:password，而是输入 domainuser:password.

We were having the same issue in our code and we fixed it by adding the domain name before the user name. Instead of entering user:password, enter domainuser:password.

希望这会有所帮助.

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