根据相同的名称分隔数组
根据相同的名称分隔数组
提问于 2020-09-11 13:30:46
我有一个具有相同名称的多个对象的数组。如何根据名称分隔数组？
我试过用地图，过滤器，但没有工作，最后我用的也减少，但没有得到输出。
const result = Object.values(this.optionsArr.reduce((a, c) => {
a[c.name] = a[c.name] || {name: c.name};
a[c.name].options.push(c.options);
return a;
}, {})).map(item => ({...item, accessDetails: Object.values(item.accessDetails)}));投入：
[{name: "audio", options:{name:'true', value: 'T'}},
{name: "audio", options:{name:'false', value: 'F'}},
{name: "audio", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'true', value: 'T'}},
{name: "video", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'false', value: 'F'}},
{name: "call", options:{name:'true', value: 'T'}},
{name: "call", options:{name:'No', value: 'N'}},
{name: "call", options:{name:'false', value: 'F'}}]预期产出
[{name: "audio", options:[{name:'true', value: 'T'},{name:'false', value: 'F'},{name:'yes', value: 'Y'}]},
{name: "video", options:[{name:'true', value: 'T'},{name:'false', value: 'F'},{name:'yes', value: 'Y'}]},
{name: "call", options:[{name:'true', value: 'T'},{name:'false', value: 'F'},{name:'yes', value: 'Y'}]}]回答 4
Stack Overflow用户
回答已采纳
发布于 2020-09-11 13:41:06
您可以使用对象来存储同名的元素，从而执行约简操作。
const arr = [{name: "audio", options:{name:'true', value: 'T'}},
{name: "audio", options:{name:'false', value: 'F'}},
{name: "audio", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'true', value: 'T'}},
{name: "video", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'false', value: 'F'}},
{name: "call", options:{name:'true', value: 'T'}},
{name: "call", options:{name:'No', value: 'N'}},
{name: "call", options:{name:'false', value: 'F'}}];
const res = Object.values(arr.reduce((acc,{name, options})=>{
acc[name] = acc[name] || {name, options: []};
acc[name].options.push(options);
return acc;
}, {}));
console.log(res);
Stack Overflow用户
发布于 2020-09-11 13:52:59
使用reduce是正确的，但是您可以减少项目并将其推到Array上，这样可以为您提供所需的输出，而不是还原为Object：
this.optionsArr.reduce((p, c) => {
if (!p.some(a => a.name === c.name)) {
p.push({ name: c.name, options: []});
}
p.find(a => a.name === c.name).options.push(c.options);
return p;
}, []);Stack Overflow用户
发布于 2020-09-11 14:02:39
您的代码几乎是正确的，您只是忘了作为一个空数组启动选项，所以推送可以工作：
将, options: []添加到还原函数中的第一个内联中
const result = Object.values(this.optionsArr.reduce((a, c) => {
a[c.name] = a[c.name] || {name: c.name, options: []};
a[c.name].options.push(c.options);
return a;
}, {})).map(item => ({...item, accessDetails: Object.values(item.accessDetails)}));页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/63848094
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