如何获得Alfresco Action状态

本文介绍了如何获得Alfresco Action状态的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有问题。我有Alfresco4。我要为特定文件夹创建一个规则（用于INBOUND，OUTBOUND，UPDATE）和操作。但是，当我将动作挂接到Java类（从ActionExecuterAbstractBase扩展）时，我不知道如何获取节点的状态（发生了什么-删除，修改或创建了节点）。解决此问题的一种方法-比较节点中的upload_date和Modifyd_date属性。但这不好。如果有人知道该怎么做，请回答。 谢谢。

I have a problem. I have Alfresco 4. I'm create a rule(for INBOUND, OUTBOUND, UPDATE) and action for my scpecific folder. But when I hook the action in my java class(extending from ActionExecuterAbstractBase) I don't know, how to get status for node(what's happen - deleted, modified or created node). One way for resolved this problem - it's a comparison uploaded_date and modified_date properties from a node. But it's not good. If anyone know how to do it, please, answer. Thank you.

推荐答案

有2种方法： 1）创建3个类，

2 ways to do this: 1) create 3 classes, which gets triggered by the right type.

2）我只想为您的操作添加一个额外的参数，例如状态。然后按照入站规则设置入站等。然后，您可以阅读action.getParameterValue（检查自定义操作Wiki ）。

2) I would just ad an extra parameter to your action, like status. And at the inbound rule just set inbound, etc. Then you can read the action.getParameterValue (check the Custom Actions Wiki).