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#1
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#2
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I have a question relating to dimension members that I hope has an easy answer that I'm just missing. Here's the situation: We have a sales cube with a dimension called "Accounts". Currently the member properties of the "Accounts" dimension are address, city, state, and zip. The problem I have is with an additional property of the accounts called "orgMembership". Each account can be a member of zero or more organizations. This is stored in the relational schema as an integer field made up of bit flags for each organization (1 = Org 1, 2 = Org 2, 4 = Org 3, etc.). I need to be able to represent this in the sales cube. I suppose I could create a dimension for each organization, but I'd like to avoid doing that if possible (there are 20+ organizations). My original thought was to create a dimension with a member for each organization, but then I couldn't figure out how to relate it to the "Accounts" dimension. Any suggestions? Thanks in advance, Sean Carpenter |
#3
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I'd be tempted to create a virtual dim from the member property. No maintenance issues and no size issue. "Sean Carpenter" wrote: I have a question relating to dimension members that I hope has an easy answer that I'm just missing. Here's the situation: We have a sales cube with a dimension called "Accounts". Currently the member properties of the "Accounts" dimension are address, city, state, and zip. The problem I have is with an additional property of the accounts called "orgMembership". Each account can be a member of zero or more organizations. This is stored in the relational schema as an integer field made up of bit flags for each organization (1 = Org 1, 2 = Org 2, 4 = Org 3, etc.). I need to be able to represent this in the sales cube. I suppose I could create a dimension for each organization, but I'd like to avoid doing that if possible (there are 20+ organizations). My original thought was to create a dimension with a member for each organization, but then I couldn't figure out how to relate it to the "Accounts" dimension. Any suggestions? Thanks in advance, Sean Carpenter |
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