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Get All Requirement Project area Name from a query.


Preetam Sikdar (61316) | asked May 14 '13, 1:43 a.m.
Hi,
The requirement is that:-
There will be 3 fields
     One User id:-  preetam@in.ibm.com
      Password:-    ***********
      RM URL:-       https://(host):(Port)/rm
      Now the user will click the GO button
So from this information we will call the OSLC and want to get all the project area where users have permission.
So we want a query from where we can get all the project area name by giving valid user id and password and the url of the rm.

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Fabian Lomeli (2213) | answered May 14 '13, 6:01 p.m.
JAZZ DEVELOPER
Not sure if this fits your needs but you can find out more information about how to get that through OSLC in this wiki:
Using OSLC capabilities in the Requirements Management application

See the entry about ...rm/discovery/RMCatalog
Preetam Sikdar selected this answer as the correct answer

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Preetam Sikdar (61316) | answered May 15 '13, 10:18 a.m.
Thanks Fabian. I have resolved the issues.
Right now i stuck in another place is that, i am creating new artifacts through OSLC in the rm, but all the artifacts is creating in the root folder.
But  i want it to create in a specific folder that present in the root folder.
So that i don't have to replace manually all the artifacts from root folder to that particular folder.
So is there any query present to create the artifacts at that particular folder.
can you provide some example.

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Fabian Lomeli (2213) | answered May 15 '13, 12:10 p.m.
JAZZ DEVELOPER
You can create the artifact in an specific folder, just need the folder resource URI to use it when creating artifact request.  In the same Using OSLC capabilities in the Requirements Management application article, in the 'Managing artifacts' -> '2. Creating artifacts' you may encounter more information.  Basically you will need to replace the
...
<nav:parent rdf:resource="https://grarrc.ibm.com:9443/rm/folders/_zpWqELrZEeG3OKR6fZ1svQ"/>
...
by your folder resource uri you want to create your artifact under.

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