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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T05:20:56+00:00 2026-05-27T05:20:56+00:00

I am thinking about starting my first CouchDB project and coming from an ORM

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I am thinking about starting my first CouchDB project and coming from an ORM background I am concerned how to create my documents that may be difficult to maintain.

For example, if I have the following model :

A *—>(1)B

which means for every A object there is a B object and there are many instances of A that can share a B object. In this case there are pointers/foreign key in A to B.

I could create a document that contains all the A data and the B data. However, the issue I have is if at a later stage (after 10000s of documents are created), I may need to change some data which means I have to update all my documents.

In an ORM/normalised database world I would simple update B and all my references are now up to database.

How do I handle this in CouchDB or is the NoSQL approach not suited for these types of situations?

JD

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  1. Editorial Team
    Editorial Team
    2026-05-27T05:20:56+00:00Added an answer on May 27, 2026 at 5:20 am

    The general answer to this question: There is no general answer to this question.

    The point is that in NoSQL, the data structure is not dictated by the data, but rather by the queries the data structure must support. So, rather than using the same pattern for each and every instance of a 1:N or M:N association problem, the NoSQL way is to use different patterns depending on your specific needs. These could be, for instance:

    • Write/Read Ratio
    • Specific database features that make embedding easier or harder
    • The types of queries you need to support
    • Performance considerations on how the data can be indexed, sharded, federated or in any other way split or cached

    Generally, my feeling is that beginners tend to ‘over-embed’, but I can speak for MongoDB only. Embedding is a powerful feature, but embedded objects are not ‘first-class citizens’, so it shouldn’t be used as a replacement for every 1:n relation. Only for some 🙂

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