Question

我想编写一个函数，其规范在下面的代码段中描述，这是我当前的实现。它确实有效。然而，我一直试图将它写成无点并完全作为ramda函数的组合而无法找到解决方案。这个问题与obj => map(key => recordSpec[key](obj[key])有关，我无法通过这种方式减少这一点，我可以将整个事情写成无点的。

我该怎么办？

/** * check that an object : * - does not have any extra properties than the expected ones (strictness) * - that its properties follow the defined specs * Note that if a property is optional, the spec must include that case * @param {Object.<String, Predicate>} recordSpec * @returns {Predicate} * @throws when recordSpec is not an object */ function isStrictRecordOf(recordSpec) { return allPass([ // 1. no extra properties, i.e. all properties in obj are in recordSpec // return true if recordSpec.keys - obj.keys is empty pipe(keys, flip(difference)(keys(recordSpec)), isEmpty), // 2. the properties in recordSpec all pass their corresponding predicate // For each key, execute the corresponding predicate in recordSpec on the // corresponding value in obj pipe(obj => map(key => recordSpec[key](obj[key]), keys(recordSpec)), all(identity)), ] ) }

例如， isStrictRecordOf({a : isNumber, b : isString})({a:1, b:'2'}) -> true isStrictRecordOf({a : isNumber, b : isString})({a:1, b:'2', c:3}) -> false isStrictRecordOf({a : isNumber, b : isString})({a:1, b:2}) -> false

Answer 1

实现这一目标的一种方法是使用R.where，它使用像recordSpec这样的规范对象，并使用第二个对象的相应键中的值应用每个谓词。

您的功能将如下所示：

const isStrictRecordOf = recordSpec => allPass([
  pipe(keys, flip(difference)(keys(recordSpec)), isEmpty),
  where(recordSpec)
])

与Ramda一起检查帮助器

1 个答案: