Conditionally Chain Methods in TypeScript

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I needed to dynamically create yup validations in TypeScript, which are accomplished by method chaining.

I only needed to chain .required in the case that requiredBool was true, as below:

const x = number()
  .typeError(`age must be a number`)
  .required(`age is required`) // Only run if requiredBool is true

The naive way to solve this that I don’t like would be:

let x = number()
  .typeError(`age must be a number`)
if (requiredBool)
  x = x.required(`age is required`) // Only runs if requiredBool is true

However, the above code does not follow the functional style and looks ugly for longer method chains, so I created an abstraction for this which you can define somewhere in your project to use:

export const chainInit = <InitialObject, TransformedObject>(value: InitialObject) => {
  return {
    if(condition: boolean, thanF: (param: InitialObject) => TransformedObject, elseF?: (param: InitialObject) => TransformedObject) {
      return chainInit(condition ? thanF(value) : elseF ? elseF(value) : value);
    },
    chain(f: (param: InitialObject) => TransformedObject) {
      return chainInit(f(value));
    },
    finish() {
      return value;
    },
  };
};

Thus, it becomes more functional and cleaner in my opinion:

chainInit(number())
  .chain(scma => scma.typeError(`age must be a number`))
  .if(requiredBool, scma => scma.required(`age is required`))
  .finish();