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