我有一个工作流,我有音频内容。我需要访问其他应用程序中的这些内容(使用JavaScript)。我用GET方法试图将此网址:如何检索Alfresco工作流程的内容?
(在这个URL,id为“1a7be6f8-0c50-4995-a211-1736642db06a”是包的工作流任务的识别器。)
但是,响应下一个XML:
<?xml version="1.0" encoding="UTF-8"?>
<feed xmlns="http://www.w3.org/2005/Atom" xmlns:app="http://www.w3.org/2007/app" xmlns:cmisra="http://docs.oasis-open.org/ns/cmis/restatom/200908/" xmlns:cmis="http://docs.oasis-open.org/ns/cmis/core/200908/" xmlns:alf="http://www.alfresco.org" xmlns:opensearch="http://a9.com/-/spec/opensearch/1.1/">
<author><name>admin</name></author>
<generator version="4.2.0 (r56674-b4848)">Alfresco (Community)</generator>
<icon>http://localhost:8086/alfresco/images/logo/AlfrescoLogo16.ico</icon>
<id>urn:uuid:1a7be6f8-0c50-4995-a211-1736642db06a-children</id>
<link rel="service" href="http://localhost:8086/alfresco/service/cmis"/>
<link rel="self" href="http://localhost:8086/alfresco/service/cmis/s/SpacesStore/i/1a7be6f8-0c50-4995-a211-1736642db06a/children?alf_ticket=TICKET_f9906d69befbc49668b92ddf372d62532a29ce7d"/>
<link rel="via" href="http://localhost:8086/alfresco/service/cmis/s/workspace:SpacesStore/i/1a7be6f8-0c50-4995-a211-1736642db06a"/>
<link rel="up" href="http://localhost:8086/alfresco/service/cmis/s/workspace:SpacesStore/i/13dd8d00-4ccd-4894-87fc-0b055cf41a4b/children" type="application/atom+xml;type=feed"/>
<link rel="down" href="http://localhost:8086/alfresco/service/cmis/s/workspace:SpacesStore/i/1a7be6f8-0c50-4995-a211-1736642db06a/descendants" type="application/cmistree+xml"/>
<link rel="http://docs.oasis-open.org/ns/cmis/link/200908/foldertree" href="http://localhost:8086/alfresco/service/cmis/s/workspace:SpacesStore/i/1a7be6f8-0c50-4995-a211-1736642db06a/tree" type="application/atom+xml;type=feed"/>
<title>1a7be6f8-0c50-4995-a211-1736642db06a Children</title>
<updated>2015-05-27T11:18:13.600-04:00</updated>
<opensearch:totalResults>0</opensearch:totalResults>
<opensearch:startIndex>0</opensearch:startIndex>
<opensearch:itemsPerPage>-1</opensearch:itemsPerPage>
<cmisra:numItems>0</cmisra:numItems>
</feed>
我不知道如何使用这个XML我的目的。我需要听到内容(mp3音频文件)并在我的自定义应用程序中修改其属性。
而且我想下一个URL(GET):
但结果是:网页脚本状态404 - 找不到
如何检索的内容工作流程?有一些RESTful URL?
感谢您的任何帮助。
问候, 巴勃罗。