const isObject = (obj) => Object.prototype.toString.call(obj) === '[object Object]';
function toArray1(children) {
const ret = children.reduce(
(acc, cur) => {
if (cur === null) return acc;
if (Array.isArray(cur)) {
acc = acc.concat(toArray1(cur));
} else if (isObject(cur) && cur.props) {
acc = acc.concat(toArray1(cur.props.children));
} else {
acc.push(cur);
}
return acc;
},
[]
);
return ret;
}
const children = [
1,
null,
[2, 3],
{
props: {
children: [4, 5]
}
}
];
const ret = toArray1(children);
console.log(ret);
回复 (7)
Y
const { pipe, path, has, both, concat, cond, T, identity, is, reduce, append, isNil, flip, useWith } = require('ramda')
const isNull = flip(isNil);
const isArray = flip(is(Array));
const isObjectAndHasProps = flip(both(is(Object), has('props')));
const toArray1 = reduce(
cond([
[isNull, identity],
[isArray, useWith(concat, [identity, t => toArray1(t)])],
[isObjectAndHasProps, useWith(concat, [identity, pipe(path(['props', 'children']), t => toArray1(t))])],
[T, flip(append)]
]),
[]
)
不可读
A
挺简单的逻辑为啥一定要用这些函数...
来个不严谨的写法🙈
const childrenStr = JSON.stringify(children);
const ret = childrenStr.match(/\d+/g) || [];
S
引用 abiuDoIT挺简单的逻辑为啥一定要用这些函数... 来个不严谨的写法🙈
@abiuDoIT 这解法 🐂 🐄 🐮
S
难道不是一句话的事吗？
children.flatMap(item => item?.props?.children || item).filter(i => i != null)
A
引用 Shonke难道不是一句话的事吗？
@Shonke ?. 现在js 支持这写法了? 😳
S
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