Java 类javax.naming.ldap.HasControls 实例源码

public Object getObjectFromNameClassPair(final NameClassPair nameClassPair) {
    if (!(nameClassPair instanceof Binding)) {
        throw new IllegalArgumentException("Parameter must be an instance of Binding");
    }

    Binding binding = (Binding) nameClassPair;
    Object object = binding.getObject();
    if (object == null) {
        throw new ObjectRetrievalException("Binding did not contain any object.");
    }
    Object result = null;
    if (nameClassPair instanceof HasControls) {
        result = mapper.mapFromContextWithControls(object, (HasControls) nameClassPair);
    }
    else {
        result = mapper.mapFromContext(object);
    }
    return result;
}

public T getObjectFromNameClassPair(final NameClassPair nameClassPair) throws NamingException{
    if (!(nameClassPair instanceof Binding)) {
        throw new IllegalArgumentException("Parameter must be an instance of Binding");
    }

    Binding binding = (Binding) nameClassPair;
    Object object = binding.getObject();
    if (object == null) {
        throw new ObjectRetrievalException("Binding did not contain any object.");
    }
    T result;
    if (nameClassPair instanceof HasControls) {
        result = mapper.mapFromContextWithControls(object, (HasControls) nameClassPair);
    }
    else {
        result = mapper.mapFromContext(object);
    }
    return result;
}

T mapFromContextWithControls(final Object ctx, final HasControls hasControls);

T mapFromContextWithControls(final Object ctx, final HasControls hasControls) throws NamingException;