javascript - Joi validator only one of keys - Stack Overflow

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I am working on an api that should allow multiple params but for three of them I would like to allow only one of them. It easier with values for each key but I am wondering if Joi allows it too or I should add extra validation logic in my server.

In short, for keys a, b or c I want to fail any request that has more than one of the three, so:

1. /?a=value is a valid request.

2. /?b=value&c=value2 is invalid.

Thanks!

I am working on an api that should allow multiple params but for three of them I would like to allow only one of them. It easier with values for each key but I am wondering if Joi allows it too or I should add extra validation logic in my server.

In short, for keys a, b or c I want to fail any request that has more than one of the three, so:

1. http://myapi./?a=value is a valid request.

2. http://myapi./?b=value&c=value2 is invalid.

Thanks!

• javascript
• node.js
• joi

Share Improve this question Follow edited Sep 2, 2019 at 5:58 DSCH asked Feb 14, 2019 at 17:02 DSCHDSCH 2,37611 gold badge2020 silver badges2929 bronze badges

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2 Answers 2

Sorted by: Reset to default 13

You're looking for object.xor(peers) if exactly one of a, b, or c is required.

Defines an exclusive relationship between a set of keys where one of them is required but not at the same time where:

• peers - the exclusive key names that must not appear together but where one of them is required. peers can be a single string value, an array of string values, or each peer provided as an argument.

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

Or, object.oxor(peers) if only one of a, b, or c is allowed, but none are required.

Defines an exclusive relationship between a set of keys where only one is allowed but none are required where:

• peers - the exclusive key names that must not appear together but where none are required.

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

I guess I will do it with a Joi.try.

const one_of_them = {
      query: Joi.alternatives().try(
        {
          a: Joi.string().required(),
          b: Joi.string().invalid().required(),
          c: Joi.string().invalid().required(),
        },
        {
          b: Joi.string().required(),
          a: Joi.string().invalid().required(),
          c: Joi.string().invalid().required(),
        },
        {
          c: Joi.string().required(),
          a: Joi.string().invalid().required(),
          b: Joi.string().invalid().required(),
        },
      ),
    };

But, maybe another solution can be better.

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