Example 55: clarify groupby in combination with relationships of instance cardinality zero

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    • Type: Bug
    • Resolution: Fixed
    • Priority: Major
    • V4.0_CSD04
    • Affects Version/s: V4.0_CS02
    • Component/s: Data Aggregation
    • None
    • Environment:

      Applied

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      Explicitly state what "left-outer-join fashion" means, i.e. not inventing dummy records with null values, instead use empty arrays that match the $expand structure:

       

      { "@odata.id": null, "Country": "France", "Sales": [ ] }

       

      Fix the unbalanced curly braces in all result rows while we are at it.

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      Explicitly state what "left-outer-join fashion" means, i.e. not inventing dummy records with null values, instead use empty arrays that match the $expand structure:   { "@odata.id" : null , "Country" : "France" , "Sales" : [ ] }   Fix the unbalanced curly braces in all result rows while we are at it.

      Example 55:

      GET ~/Customers?$apply=groupby((Country,Sales/Product/Name))

      The example result does not contain an entry for country France. The example data contains a customer in France that does not have any sales.

      The prose text before the example states expansion "in a left-outer-join fashion".

      This would suggest that a result row is missing:

      { "@odata.id": null, "Country": "France", "Sales": [ { "Product":

      { "Name": null }

      } ] }

      This would be consistent with both the left-outer-join statement and the URL conventions for path expressions where properties of related entities are treated as null if no entity is related:

      • the left-outer join would produce a single Sales row containing only null values, including the related product and its name
      • this would create a ("France",null) group during aggregation
      • folding back into the original shape would create the additional entry

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            Reporter:
            handl
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              Created:
              Updated:
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