我想要实现的基本上是这样的:
R.mergeDeepRight(
{ age: 40, contact: { email: '[email protected]' }},
{ name: 'fred', age: 10, contact: { email: '[email protected]' }}
);
但在结果对象中没有{ name: 'fred' }
。
只应应用第一个对象中的键。
我会把mergeDeepRight
,pick
和keys
组合成一个可重用的函数,如下所示:
const funkyMerge = (o1, o2) =>
mergeDeepRight(o1, pick(keys(o1), o2))
console.log(funkyMerge(
{ age: 40, contact: { email: '[email protected]' }},
{ name: 'fred', age: 10, contact: { email: '[email protected]' }}
))
<script src="https://bundle.run/[email protected]"></script><script>
const {mergeDeepRight, pick, keys} = ramda </script>
您可以在omit之后使用mergeDeepRight来省略不需要的密钥
let obj1 = { age: 40, contact: { email: '[email protected]' }}
let obj2 = { name: 'fred', age: 10, contact: { email: '[email protected]' }}
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>
这里ommitedKeys变量包含Object1中不存在的所有键,
替代是使用Pick
let obj1 = { age: 40, contact: { email: '[email protected]' }}
let obj2 = { name: 'fred', age: 10, contact: { email: '[email protected]' }}
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>