如何将对象数组转换为在打字稿中具有动态键的单个对象
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【中文标题】如何将对象数组转换为在打字稿中具有动态键的单个对象【英文标题】：How to convert array of objects to single object which has dynamic key in typescript 【发布时间】：2019-09-18 17:01:18 【问题描述】：这个问题可能与常见问题类似，但这个问题有一些不同的方法。
在我的 Angular 7 应用程序中，我有以下 5 个数组，需要根据 id 使用动态键将其转换为下面的单个对象。
"enabled-41": true,
"enabled-42": true,
"enabled-43": true,
"enabled-44": true,
"enabled-45": false,
"abc-41": "some description 1",
"abc-42": "some description 12",
"abc-43": "some description 123",
"abc-44": "some description 1234",
"abc-45": null,
"def-41": "some description 2",
"def-42": "some description 23",
"def-43": "some description 234",
"def-44": "some description 2345",
"def-45": null,
"type-41": "def",
"type-42": "abc",
"type-43": "def",
"type-44": "abc",
"type-45": null,
"weight-41": "25",
"weight-42": "25",
"weight-43": "25",
"weight-44": "25",
"weight-45": null
let arr = [
"id": 41,
"abc": "some description 1",
"def": "some description 2",
"type": "def",
"Criteria":
"id": 5,
"question": "follow-up",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 42,
"abc": "some description 12",
"def": "some description 23",
"type": "abc",
"Criteria":
"id": 1,
"question": "coverage",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 43,
"abc": "some description 123",
"def": "some description 234",
"type": "def",
"Criteria":
"id": 4,
"question": "Price",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 44,
"abc": "some description 1234",
"def": "some description 2345",
"type": "abc",
"Criteria":
"id": 3,
"question": "Exchange",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 45,
"Criteria":
"id": 2,
"definition": "definition conent",
"question": "Random",
"status": true
,
"type": null,
"abc": null,
"def": null,
"weight": 0,
"enabled": false
];
let result = arr.reduce(function(obj, item)
obj[item] = item.value;
return obj;
, )
console.log(result);我尝试过使用 reduce 函数，但无法根据动态键（用连字符连接 id）获得正确的方法来转换为上述格式的单个对象。
有人可以帮我解决这个问题吗？
【问题讨论】：
Criteria 对象呢？应该发生什么？它应该被丢弃吗？
是Criteria对象可以被丢弃，想得到同上的单个对象
【参考方案1】：
您的代码几乎就在那里。但是不能保证对象键的顺序。在reduce回调函数内部添加累加器中的键和对应的值。
在创建对象键时使用模板文字和方符号
let arr = [
"id": 41,
"abc": "some description 1",
"def": "some description 2",
"type": "def",
"Criteria":
"id": 5,
"question": "follow-up",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 42,
"abc": "some description 12",
"def": "some description 23",
"type": "abc",
"Criteria":
"id": 1,
"question": "coverage",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 43,
"abc": "some description 123",
"def": "some description 234",
"type": "def",
"Criteria":
"id": 4,
"question": "Price",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 44,
"abc": "some description 1234",
"def": "some description 2345",
"type": "abc",
"Criteria":
"id": 3,
"question": "Exchange",
"definition": "definition content",
"status": true
,
"weight": 25,
"enabled": true
,
"id": 45,
"Criteria":
"id": 2,
"definition": "definition conent",
"question": "Random",
"status": true
,
"type": null,
"abc": null,
"def": null,
"weight": 0,
"enabled": false
];
let result = arr.reduce(function(obj, item)
obj[`enabled-$item.id`] = item.enabled;
obj[`abc-$item.id`] = item.abc;
obj[`def-$item.id`] = item.def;
obj[`type-$item.id`] = item.type;
obj[`weight-$item.id`] = item.weight;
return obj;
, );
console.log(result)【讨论】：
谢谢，对象键顺序有可能吗？ @UI_Dev 希望此链接***.com/questions/5525795/… 对您有所帮助 @UI_Dev 有人做过【参考方案2】：您可以将reduce 与Object.keys 一起使用，并将您希望排除的所有键放在一个数组中并检查：
let arr = ["id":41,"abc":"some description 1","def":"some description 2","type":"def","Criteria":"id":5,"question":"follow-up","definition":"definition content","status":true,"weight":25,"enabled":true,"id":42,"abc":"some description 12","def":"some description 23","type":"abc","Criteria":"id":1,"question":"coverage","definition":"definition content","status":true,"weight":25,"enabled":true,"id":43,"abc":"some description 123","def":"some description 234","type":"def","Criteria":"id":4,"question":"Price","definition":"definition content","status":true,"weight":25,"enabled":true,"id":44,"abc":"some description 1234","def":"some description 2345","type":"abc","Criteria":"id":3,"question":"Exchange","definition":"definition content","status":true,"weight":25,"enabled":true,"id":45,"Criteria":"id":2,"definition":"definition conent","question":"Random","status":true,"type":null,"abc":null,"def":null,"weight":0,"enabled":false];
let exclude = ["id", "Criteria"];
let result = arr.reduce((acc, curr) =>
let id = curr.id;
Object.entries(curr).forEach(([k, v]) =>
if (!exclude.includes(k)) acc[`$k-$id`] = v;
);
return acc;
, );
console.log(result);【讨论】：
【参考方案3】：假设您要排除所有值为 object 的属性，也许您可以采用这个通用的想法，即使用 Object.entries() 遍历内部对象和一些 destructuring 特征。
let arr=["id":41,"abc":"some description 1","def":"some description 2","type":"def","Criteria":"id":5,"question":"follow-up","definition":"definition content","status":true,"weight":25,"enabled":true,"id":42,"abc":"some description 12","def":"some description 23","type":"abc","Criteria":"id":1,"question":"coverage","definition":"definition content","status":true,"weight":25,"enabled":true,"id":43,"abc":"some description 123","def":"some description 234","type":"def","Criteria":"id":4,"question":"Price","definition":"definition content","status":true,"weight":25,"enabled":true,"id":44,"abc":"some description 1234","def":"some description 2345","type":"abc","Criteria":"id":3,"question":"Exchange","definition":"definition content","status":true,"weight":25,"enabled":true,"id":45,"Criteria":"id":2,"definition":"definition conent","question":"Random","status":true,"type":null,"abc":null,"def":null,"weight":0,"enabled":false];
let result = arr.reduce((obj, id, ...rest) =>
Object.entries(rest).forEach(([k, v]) =>
if (Object(v) !== v) obj[`$k-$id`] = v;
);
return obj;
, );
console.log(result);.as-console background-color:black !important; color:lime;
.as-console-wrapper max-height:100% !important; top:0;【讨论】：
【参考方案4】：哦，伙计……我刚刚被打败了。这是我的解决方案。
let arr= [] // hold the final object array
let keys = [] // temp item to hold the value of each key
// iterate over each key
Object.keys(input).forEach((key) =>
let pair = key.split('-') // split the key into the real key and the index
// if the index isn't in the array, push it there (this keeps the same order)
if (keys.indexOf(pair[1])===-1)
keys.push(pair[1])
// use object.assign to add the keys to the existing object in the right place in the array.
arr[keys.indexOf(pair[1])] = Object.assign(, arr[keys.indexOf(pair[1])], [pair[0]]: input[key], id: pair[1] )
)
【讨论】：
【参考方案5】：function getFilteredData(arr)
const result = ;
arr.forEach(item =>
const id, Criteria, ...rest = item;
Object.entries(rest).map(([key, value]) =>
result[`$key-$id`] = value;
);
);
return result;
let arr = [
id: 41,
abc: 'some description 1',
def: 'some description 2',
type: 'def',
Criteria:
id: 5,
question: 'follow-up',
definition: 'definition content',
status: true
,
weight: 25,
enabled: true
,
id: 42,
abc: 'some description 12',
def: 'some description 23',
type: 'abc',
Criteria:
id: 1,
question: 'coverage',
definition: 'definition content',
status: true
,
weight: 25,
enabled: true
,
id: 43,
abc: 'some description 123',
def: 'some description 234',
type: 'def',
Criteria:
id: 4,
question: 'Price',
definition: 'definition content',
status: true
,
weight: 25,
enabled: true
,
id: 44,
abc: 'some description 1234',
def: 'some description 2345',
type: 'abc',
Criteria:
id: 3,
question: 'Exchange',
definition: 'definition content',
status: true
,
weight: 25,
enabled: true
,
id: 45,
Criteria:
id: 2,
definition: 'definition conent',
question: 'Random',
status: true
,
type: null,
abc: null,
def: null,
weight: 0,
enabled: false
];
console.log(getFilteredData(arr));【讨论】：
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