Require exactly two of three fields to be non-empty

[amaury1093]

Context

• node version: Not relevant
• joi version: 13.1.1
• environment (node, browser): Not relevant
• used with (hapi, standalone, ...): Not relevant
• any other relevant information:

What are you trying to achieve or the steps to reproduce ?

const schema = Joi.object().keys({
    a: Joi.string(),
    b: Joi.string(),
    c: Joi.string()
});

I want a, b, c to be exactly 2 out of 3 non-empty. I.e.:

• if a, b are not empty, c should not be set
• idem with circular permutation of [a,b,c]
• of course, if 2 or more are empty, throw an error too

Which result you had ?

A StackOverflow answer led to me this solution:

Joi.alternatives().try(
    Joi.object().keys({
        a: Joi.string().required(),
        b: Joi.string().required(),
    }),
    Joi.object().keys({
        b: Joi.string().required(),
        c: Joi.string().required()
    }),
    Joi.object().keys({
        a: Joi.string().required(),
        c: Joi.string().required()
    })
);

What did you expect ?

A more scalable solution to this. If I wanted a 2-of-4 non-empty I'd have to write 6 Joi alternatives.

[DavidTPate]

This is an interesting one. I don't think we have an exact way to say a and b or a and c or b and c etc. without using Joi.alternatives().

This technically solves your current test, let us know if it works for you:

const schema = Joi.object().keys({
    a: Joi.string(),
    b: Joi.string(),
    c: Joi.string()
}).min(2).max(2);

schema.validate({ a: '1', c: '2' })

So the above is stating that you must have exactly 2 keys present and they must be a, b, or c.

[amaury1093]

I really like your solution, however it does not solve my problem, as I have, d, e... and other optional and non-optional fields.

const schema = Joi.object().keys({
    a: Joi.string(),
    b: Joi.string(),
    c: Joi.string(),
    d: Joi.number(),
    // ... other fields
})

And only [a,b,c] need to have the 2-out-of-3 rule.

[Marsup]

It shouldn't be too difficult with an extension, but I don't see a way to do it ootb.

[lock]

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