Subscription owner and multiple subscriptions

  • We are using reporting service 2019 Enterprise.

    There is an issue with running the subscription the other people created. For example two user are both content manager role.

    And the first user created the subscription, and she becomes the owner of the subscription automatically.  She runs the report with no problem. But when the second user runs the report it failed.

    But when the second user create another subscription exactly like the first user's, and he can run the report without problem.

    The only difference is the owner of the subscription are different.

    We don't want to go the route that each user that runs the subscription has to create a subscription.

    How to resolve this? Thanks

  • Never mind. Figured out. I found I cannot delete my own post

    • This reply was modified 3 years, 9 months ago by  sqlfriend.
  • sqlfriend wrote:

    Never mind. Figured out. I found I cannot delete my own post

    So what was the fix you figured out?

    --Jeff Moden


    RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
    First step towards the paradigm shift of writing Set Based code:
    ________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.

    Change is inevitable... Change for the better is not.


    Helpful Links:
    How to post code problems
    How to Post Performance Problems
    Create a Tally Function (fnTally)

  • The rsExecRole  needs to be granted to the user who run the subscription

  • K.  Thanks.

    --Jeff Moden


    RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
    First step towards the paradigm shift of writing Set Based code:
    ________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.

    Change is inevitable... Change for the better is not.


    Helpful Links:
    How to post code problems
    How to Post Performance Problems
    Create a Tally Function (fnTally)

  • This was removed by the editor as SPAM

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