3

I tried calling a 2010 workflow from a 2013 workflow and it does not work. Everything I’ve read indicates that its straight forward. From the SPD:

Start 2010Workflow with no parameters on Current Item (Output to Data source)

The 2013 workflow never makes it past this point. As a test the ‘2010Workflow’ just logs to the workflow history. Workflow error indicates this:

Details: System.ApplicationException: HTTP 400 {"error":{"code":"-2147024809, System.ArgumentException","message":{"lang":"en-US","value":"associationName"}}}

I’ve read a little about using the api and HttpSend method. It’s a bit more involved and almost breaks my OOTB mandate… but if it works then so be it. Any recommendations on how best to proceed?

Kind regards,

Improve this question
3
  • More info - I have two WF. A 2013 WF, associated to a list, which executes on change. It calls a 2010 WF that should create a new DocSet. When the 2010 WF is associated to the DocSet, manually running the 2010 WF successfully creates the DocSet. However the 2013 WF will not trigger the 2010 WF. When a 2010 WF is associated to the List, manually running the 2010 WF does not create the DocSet. However the 2013 WF will trigger the 2010 WF. Its close, but still broken.
    – MaxK-2773298
    Commented Oct 26, 2016 at 22:12
  • In 2010 workflow, instead of creating the docset, try to just log some text. If it works then we can deduce that docset creation is issue else we need to check in detail.
    – Monica Jagani
    Commented Oct 27, 2016 at 5:21
  • That has been my testing, using text logging. When 2013 WF calls 2010 WF (associated with DocSet), there is no logging. Its as if WF methods cannot leave the list they are in. I can't call a WF outside the list, and can't have the WF inside the list create a DocSet. Permissions? Settings?,.. bah!.
    – MaxK-2773298
    Commented Oct 27, 2016 at 13:54

1 Answer 1

Reset to default
0

Can you try to call your 2010 workflow under App step. If any permission related issue then app step may solve it.

Improve this answer
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.