Custom Grid query that shows Feature + User stories with particular milesonte/release only

[Anon Anon]

How can write a Custom Grid query to include only user stories that have a particular milestone or release associated. Our features span multiple releases so I want a way to see only those user stories that are scheduled for a release.

[KenLaFrance1340184]

To clarify - are you looking to see stories whose features are associated with a given release or milestone OR are you looking to show features whose stories are part of a release or milestone?

[John_Streeter]

EBSCO is all over this community I tell ya what! @Eric Nash :)

[Anon Anon]

Let me give you an example, lets say I have Feature foo that has story1 with release r1 and story 2 with release r2. I want to see a custom grid result where I want to see all Features that have user stories targeted for r1. Feature foo should be displayed with only story1 in this case.

[KenLaFrance1340184]

Simply set your custom list to view features and then put in the following query (but with the appropriately named release:  (UserStories.Release.Name = "CY2016 Q2").  This will give you all the features that have stories that are part of release 'CY2016 Q2'. Hope this helps

[Anon Anon]

thank you for your response, it is still showing Stories with different releases :(

[EricNash]

@Namrata Dubey, using this query you'll get every feature that has a user story in the specified release. If feature FOO has one story in R1 and one story in R2 and you condition a query to return features with user stories in R1 then feature FOO meets the filter criteria. Remember, in this scenario you're filtering out features, not stories. A different approach might be to query user stories instead of features (Release.Name = "R1"), and include the Feature column in the query results, ordered by Feature. You may have duplicates if a feature has two stories in R1 but it shouldn't be difficult to see them, and you can always export your custom list to CSV and de-dupe in Excel. By the way, @Ken LaFrance is my boss so in no way should this be construed as contradicting his answer ;-)