share - 自定义 Surf 平台根范围 API

Question

我想专门定制Surf Platform Root-Scoped API用户对象。这意味着向用户对象添加新属性或方法以检查用户是否在header.inc.ftl[in share] 中的某个组中，例如 `<#if user.isAdmin>

1. 我该如何实施？
2. Alfresco Root Scoped Objects可以用作 Surf Platform Root-Scoped 对象吗？

我不知道自定义冲浪平台根对象。谁能帮我？？？

Answer 1

Not quite sure what you are trying to accomplish, but the role security model is hardcoded in spring-surf/spring webscripts. There is guest, user and admin. If what you want is another analogous role you'll have to hack the spring-surf libaries, namely:

org/springframework/extensions/surf/mvc/PageView.java
org/springframework/extensions/webscripts/ScriptUser.java
org/springframework/extensions/webscripts/Description.java
org/springframework/extensions/webscripts/connector/User.java

This is what I had to do to implement user.isEmployee. This approach allows you to literally treat your new role just as the others.

you can use

<authentication>employee</authentication>

in page descriptors or

<item type="link" permission="employee" id="people">/people-finder</item>

on the navigation.

Just checking whether the user is in a certain group in a certain webscript is a whole diffrent story and does not provide the same functionality.

If what you want is the latter, you should make a call to

/alfresco/service/api/groups/{shortName}

miss and works through the response.

Update: The item permission attribute requires a little more tweaking. In header.get.js, propagate the new role to it gets processed properly in header.inc.ftl:

model.permissions =
{
    guest: user.isGuest,
    admin: user.isAdmin,
    employee : user.isEmployee
};

Answer 2

你可以试试（在JavaScript中我管理过类似的东西）：

user = Application.getCurrentUser(context);
       String userName = user.getUserName();
       user.isAdmin() >>> result return true if user logining is admin

或在 JSP 中：

#{NavigationBean.currentUser.admin == true}

抱歉，我现在注意到您在谈论 Surf Platform 根对象，但是您放在那里的链接对于 3.3 以上的 Alfresco 版本已弃用。你还在用这么老的东西？如果您设法使用 JavaScript API，您可以使用带有布尔 isAdmin() 的“人”根对象。