get nodeRef associated to Content

[soeursourire]

Hi!

Is there a fast way to get the NodeRef associated to a file contained into the repository: I have a xml file stored in the contents of admin, I know the id of this node file and I would like to get the NodeRef. With PersonServiceImpl we can do that: NodeRef getPerson(String name), is their a function NodeRef getContent(String id) somewhere in Alfresco?

Thanks

[soeursourire]

I guess the simplest way was to do this:

public NodeRef getRecordNodeRef(String recordId)
   {
      NodeRef rootNode = this.nodeService.getRootNode(storeRef);
       QueryParameterDefinition[] defs = new QueryParameterDefinition[1];
       DataTypeDefinition typeText = dictionaryService.getDataType(DataTypeDefinition.TEXT);
       defs[0] = new QueryParameterDefImpl(QName.createQName("sys", "var", namespacePrefixResolver), typeText, true,
               recordId);
       List<NodeRef> results = searchService.selectNodes(rootNode, ACTIVERECORDS_FOLDER
               + "/cm:Record.xml[@sys:node-uuid = $sys:var ]", defs, namespacePrefixResolver, false);
       if (results.size() != 1)
       {
           return null;
       }
       return results.get(0);
   }

[anweber]

Hi,

   Following the documentation, predicates are not supported in XPATH queries.

  Try this :

   public static NodeRef getNodeByUUID(ServiceRegistry serviceRegistry, StoreRef storeRef, String uuId) throws Exception{
       
       SearchService searchService = serviceRegistry.getSearchService();
        String searchString = "@sys\\:node-uuid:\"" + uuId + "\"";
         ResultSet rs = searchService.query(storeRef,SearchService.LANGUAGE_LUCENE, searchString );
        if (rs==null){
           return(null);
        }
        int nbRecords = rs.length();
        NodeRef nodeRef;
        if (nbRecords < 1) {
           nodeRef=null;
        }
        else nodeRef = rs.getNodeRef(0);
        rs.close();
        return(nodeRef);
    }

‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

   I hope this may help you.

    Regards,

             Andre