Question

我想要实现的基本上是这样的：

R.mergeDeepRight(
  { age: 40, contact: { email: 'baa@example.com' }},
  { name: 'fred', age: 10, contact: { email: 'moo@example.com' }}
);

但结果对象中没有{ name: 'fred' }。

仅应应用第一个对象中的键。

Answer 1

您可以在omit之后使用mergeDeepRight来省略不需要的键

let obj1 = { age: 40, contact: { email: 'baa@example.com' }}
let obj2 = { name: 'fred', age: 10, contact: { email: 'moo@example.com' }}

let ommitKeys = Object.keys(obj2).filter(key=> !obj1[key])

let concatValues = (k, l, r) =>  k == 'values' ? R.concat(l, r) : r

let output = R.omit(ommitKeys, R.mergeDeepRight(concatValues, obj1, obj2,))


console.log(output)

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>

Demo

这里ommitedKeys变量保存了Object1中不存在的所有键，

替代方法是使用Pick

let obj1 = { age: 40, contact: { email: 'baa@example.com' }}
let obj2 = { name: 'fred', age: 10, contact: { email: 'moo@example.com' }}

let desiredKeys = Object.keys(obj1)

let concatValues = (k, l, r) =>  k == 'values' ? R.concat(l, r) : r

let output = R.pick(desiredKeys, R.mergeDeepRight(concatValues, obj1, obj2,))


console.log(output)

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>

Answer 2

我将结合使用mergeDeepRight，pick和keys这样的可重用函数，如下所示：

const funkyMerge = (o1, o2) =>   
  mergeDeepRight(o1, pick(keys(o1), o2))

console.log(funkyMerge(
  { age: 40, contact: { email: 'baa@example.com' }},
  { name: 'fred', age: 10, contact: { email: 'moo@example.com' }}
))

<script src="https://bundle.run/ramda@0.26.1"></script><script>
const {mergeDeepRight, pick, keys} = ramda      </script>

同步两个对象并应用新值

2 个答案: