What's the best way to get a node by full path?

LLNode has a wonderful NodePath() to return the path of a node as "A:B:C". Is there a function to do the opposite - take "A:B:C" and return a node or nodeID?

I need to be able to get the matching nodeID on another server based on the path. I don't want to re-invent the wheel if there is already a way to do this.

Thanks!

Bobby

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Appu_Nair1

Have a look in webdoc.:DocRequestHandler:View for DAPI.GetNode.You can see a similar function CWS they wrote it and you can use that implemnation as well. uses in DocManService.You can use oscript to call webservices within code.

AFA how it goes when you ask livelink this qn getpathof thisnodebyid(12345) it is easy for that qn to be answered because there is a unique node sitting somewhere and all it has to do is trace the hierarchy traversing volumes(-1) when needed.

However if you ask the qn GetNodeBypath('A:B:C') it will be very expensive to tell that so you have to ask it by saying I mean business by GetNodeBypath("Start at this nodeid",'A:B:C').There could be millions of objects in a livelink having the same path.

When trying to migrate code between livelink servers one saves a dataid and say this will change to this when it goes to Prod and try to enumerate like paths between systems.You could even use Enterp rise in a config file but I have seen code fail because Enterprise can be renamed.You can create your own volume object so you will get a starting place if you want to go that route s well

PS :Livelink never had a real "transport" system.So starting with 10 they now include a 'GUID' in dtree.Somebody in OT once told me a transport system is getting readied to surface in 2014 and it may use that for something.I could be wrong.In the interim if you do not have smart tools form GCI you always do a system mapping with dataids' to get your job done

Greg_Griffiths

If you are doing this in OScript, then you can use DAPI.ListSubNodes with the query parameter being the name of the next folder, so something like :

1. Get A as a DAPI Node

2. DAPI.ListSubNodes( A, "name = 'B')

3. Assuming no duplicate names, you can extend the Query parameter to include a subtype check etc, then you just repeat step 2 for each folder in the tree.

Bobby_Brimm

Thanks, I'll take another look there. If this becomes too time consuming, I may cheat and strip the first breadcrumb out and do a hard coded transformation ( Enterprise = rootid = 2000, Content Server Categories = rootCatid = 2004 ) and use the easy function ( since it is unlikely we will change these names from instance to instance in the same company ).

Thanks!

Bobby

Bobby_Brimm

Found my solution: Searched other posts about finding DataIDs for volumens and found SQL to pull the root node by the breadcrumb name.

So to get the node by path "A:B:C" I strip off the first breadcrumb (A) and do the following select

SELECT DATAID as root , NAME FROM DTREE WHERE DATAID = - OWNERID AND NAME = 'A'

then I can simply use the DAPI.GetNode ( "B:C", root ).

I will probably make this a utility call in a module because I can see the use for this again and again.

Thanks one and all for all your thoughtful contributions!

Bobby

Appu_Nair1

Again don’t assume anything. Our enterprise is not 2000 our cat volume is not 2004 you have oscript functions that tell you what the category volume is and enterprise volume is. You never have to rely on hard codes in livelink.

From: eLink Entry: Content Server Development Forum [mailto:development@elinkkc.opentext.com]
Sent: Monday, July 01, 2013 6:45 AM
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Subject: [EXTERNAL]What's the best way to get a node by full path? 2

What's the best way to get a node by full path? 2

Posted by bjbrimm@tva.gov (Brimm, Bobby) On 07-01-2013 07:41

Thanks!

Bobby

[To post a comment, use the normal reply function]
Topic:	What's the best way to get a node by full path?
Forum:	Content Server Development Forum
Content Server:	Knowledge Center

Hans Stoop

I wouldn’t do it like that.

I would define a mapping table in code. When the first name is EWS then it is the Enterprise Workspace and I would get the ID by

Select DataID as root, Name from DTree where SubType = 141

In case you have a case sensitive SQLserver database you need to write the table name correctly.

In a similar way you can define the Categories volume.

With your current query you can also get a project returned.

Hans

From: eLink Entry: Content Server Development Forum [mailto:development@elinkkc.opentext.com]
Sent: Montag, 1. Juli 2013 14:13
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Subject: What's the best way to get a node by full path? 2

What's the best way to get a node by full path? 2

Posted bybjbrimm@tva.gov (Brimm, Bobby) On 07-01-2013 08:10

Found my solution: Searched other posts about finding DataIDs for volumens and found SQL to pull the root node by the breadcrumb name.

So to get the node by path "A:B:C" I strip off the first breadcrumb (A) and do the following select

SELECT DATAID as root,NAMEFROM DTREE WHERE DATAID =- OWNERID ANDNAME = 'A'

then I can simply use the DAPI.GetNode ( "B:C", root ).

I will probably make this a utility call in a module because I can see the use for this again and again.

Thanks one and all for all your thoughtful contributions!

Bobby

[To post a comment, use the normal reply function]
Topic:	What's the best way to get a node by full path?
Forum:	Content Server Development Forum
Content Server:	Knowledge Center

Bobby_Brimm

Subtype = 141 only gets me the Enterprise root. I need to be able to get all sorts of "root folders". Personal workspace, Categories, etc.... I checked all rows in the database and specifying dataID = -owner coupled with providing NAME seems to work for all 400+ rows in the database. If it bites me in the end, I'll know I was warned.....

Bobby_Brimm

Yes, I agree... I hate assumptions and hard coding. I found a SQL answer to the problem for the "Root" of the full path that seems to be holding up and performing nicely. You just gotta find out where to look.