创建一个对象 i 一个基于字符串值数组的对象

Question

我需要根据字符串值的数组更新对象名称，最后一个字符串值应该是数组。

我使用 array.forEach 循环，但我不知道如何在对象中找到对象(如果存在并且 myArray 包含大约 10,000 个字符串(。

const myArray = [
'/unit/unit/225/unit-225.pdf',
'/nit/nit-dep/4.11/nit-4.11.pdf',
'/nit/nit-dep/4.12/nit-4.12.pdf',
'/org/viti/viti-engine/5.1/viti-engine-5.1.pdf',
'/org/viti/viti-spring/5.1/viti-spring-5.1.pdf'
];
var parentObject = {}
myArray.forEach(res => {
res = res.slice(1, res.length);
var array = res.split("/");
array.forEach((e, i) => {
........ // here I am confused 
});
})

最终输出应为

parentObject = {
'unit': {
'unit': {
'225': {
'unit-225.pdf': []
}
}
},
'nit': {
'nit-dep': {
'4.11': {
'nit-4.11.pdf': []
},
'4.12': {
'nit-4.12.pdf': []
}
}
},
'org': {
'viti': {
'viti-engine': {
'5.1': {
'viti-engine-5.1.pdf': []
}
},
'viti-spring': {
'5.2': {
'viti-engine-5.2.pdf': []
}
}
}
}
}

Answer 1

用斜杠拆分后，使用reduce迭代到嵌套对象，如有必要，首先创建每个嵌套属性，然后将数组分配给 filename 属性：

const myArray = [
'/unit/unit/225/unit-225.pdf',
'/nit/nit-dep/4.11/nit-4.11.pdf',
'/nit/nit-dep/4.12/nit-4.12.pdf',
'/org/viti/viti-engine/5.1/viti-engine-5.1.pdf',
'/org/viti/viti-spring/5.1/viti-spring-5.1.pdf'
];
var parentObject = {}
myArray.forEach((str) => {
const props = str.slice(1).split('/');
const filename = props.pop();
const lastObj = props.reduce((a, prop) => {
if (!a[prop]) {
a[prop] = {};
}
return a[prop];
}, parentObject);
lastObj[filename] = [];
});
console.log(parentObject);

Answer 2

您也可以减少数组并减少路径。最后分配数组。

const
array = ['/unit/unit/225/unit-225.pdf', '/nit/nit-dep/4.11/nit-4.11.pdf', '/nit/nit-dep/4.12/nit-4.12.pdf', '/org/viti/viti-engine/5.1/viti-engine-5.1.pdf', '/org/viti/viti-spring/5.1/viti-spring-5.1.pdf'],
result = array.reduce((r, path) => {
var keys = path.split(///).slice(1),
last = keys.pop();
keys.reduce((o, k) => o[k] = o[k] || {}, r)[last] = [];
return r;
}, {});
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

稍微快一点的方法。

const
array = ['/unit/unit/225/unit-225.pdf', '/nit/nit-dep/4.11/nit-4.11.pdf', '/nit/nit-dep/4.12/nit-4.12.pdf', '/org/viti/viti-engine/5.1/viti-engine-5.1.pdf', '/org/viti/viti-spring/5.1/viti-spring-5.1.pdf'],
result = {};
for (let path of array) {
let keys = path.split(///).slice(1),
last = keys.pop(),
temp = result;
for (let key of keys) {
temp[key] = temp[key] || {};
temp = temp[key];
}
temp[last] = [];
}
console.log(result);

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Answer 3

您也可以采用递归方法。

只需继续移动拆分路径，直到获得每个分支的最终分配。

const myArray = [
'/unit/unit/225/unit-225.pdf',
'/nit/nit-dep/4.11/nit-4.11.pdf',
'/nit/nit-dep/4.12/nit-4.12.pdf',
'/org/viti/viti-engine/5.1/viti-engine-5.1.pdf',
'/org/viti/viti-spring/5.1/viti-spring-5.1.pdf'
];
console.log(buildTree(myArray));
function buildTree(list=[]) {
return list.reduce((node, item) => buildBranches(node, item.split(///g).filter(x => x !== '')), {});
}
function buildBranches(node={}, rest=[]) {
let key = rest.shift();
node[key] = rest.length < 2 ? { [rest.shift()] : [] } /** or rest.shift() */ : node[key] || {};
if (rest.length > 1) buildBranches(node[key], rest);
return node;
}

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