JS中的Struts2属性

Question

如何在JS中访问S2属性。我尝试了以下代码。在这里，我需要在JS函数中动态更改span内容。

在JS中：

template <typename Type>
void LinkedList<Type>::insert(const Type &item, int position){
    if(head == NULL){
        ListElement *newElement = new ListElement(item, NULL);
        newElement->next = NULL;
        head = newElement;
        cursor = head;
    }else if(position == 0){
        if(head->next == NULL){
            cursor = head;
        }
        //Create a new ListElement after the current.
        ListElement *newElement = new ListElement(item, NULL);
        //Preserve link (if there is one)
        if(cursor->next != NULL){
            newElement->next = cursor->next;
        }
        cursor->next = newElement;
        cursor = newElement;
    }else if(position == -1){
        //Add a new element before the current.
        ListElement *newElement = new ListElement(item, NULL);
        if(head->next == NULL || (cursor == head && cursor != NULL)){
            //Special case. There is only one node.  Must update head pointer.
            newElement->next = head;
            head = newElement;
            cursor = newElement;
        }else{
            //Node will be put between two other nodes.
            ListElement *iterator = head;
            while(iterator->next != cursor){
                iterator = iterator->next;
            }
            newElement->next = iterator->next;
            iterator->next = newElement;
            cursor = newElement;
        }
    }//end if line 24
}

在JSP中：

else if(position == 0){
            if(head->next == NULL){
                cursor = head;
            }
            //Create a new ListElement after the current.
            ListElement *newElement = new ListElement(item, NULL);
            //Preserve link (if there is one)
            if(cursor->next != NULL){ //LINE 71 LINE 71 LINE 71 LINE 71
                newElement->next = cursor->next;
            }

但没有任何东西可以展示。任何其他解决方案。