javascript - 将重复出现的 JSON 名称转换为对象？

Question

我有以下字符串...

{
    "Qualifications[0].DegreeType": "Degree Type 1",
    "Qualifications[0].Classification": "Clasification 1",
    "Qualifications[0].CourseTitle": "Course Title 1",
    "Qualifications[0].YearOfAward": "2018",
    "Qualifications[0].AwardingInstitution": "Awarding institution 1",
    "Qualifications[1].DegreeType": "Degree Type 2",
    "Qualifications[1].Classification": "Classification 2",
    "Qualifications[1].CourseTitle": "Course Title 2",
    "Qualifications[1].YearOfAward": "2019",
    "Qualifications[1].AwardingInstitution": "Awarding Institution 2",
    ... and so on...
 }

有没有人知道将其转换为 javascript 对象的简单方法，这样我就可以有效地...

$(Qualifications).each(i, v){
    //do stuff on whatever property of Qualifications[i]
}

我能想到的唯一方法是使用字符串操作和类似下面的东西，但它需要大量自定义代码检查每个属性的映射，我正在寻找更通用和简单的东西......

    let i = 0;
    while (Qualifications[" + i +"].whateverProperty..) {
        //add to object array
        i++;
    }

Answer 1

您可以拆分键并通过行走给定对象来减少路径。

如果不存在对象，则使用名称或数组创建一个新属性。稍后分配值。

function setValue(object, path, value) {
    var way = path.replace(/\[/g, '.').replace(/\]/g, '').split('.'),
        last = way.pop();

    way.reduce(function (o, k, i, kk) {
        return o[k] = o[k] || (isFinite(i + 1 in kk ? kk[i + 1] : last) ? [] : {});
    }, object)[last] = value;
}

var object = {
        "Qualifications[0].DegreeType": "Degree Type 1",
        "Qualifications[0].Classification": "Clasification 1",
        "Qualifications[0].CourseTitle": "Course Title 1",
        "Qualifications[0].YearOfAward": "2018",
        "Qualifications[0].AwardingInstitution": "Awarding institution 1",
        "Qualifications[1].DegreeType": "Degree Type 2",
        "Qualifications[1].Classification": "Classification 2",
        "Qualifications[1].CourseTitle": "Course Title 2",
        "Qualifications[1].YearOfAward": "2019",
        "Qualifications[1].AwardingInstitution": "Awarding Institution 2"
    },
    result = {};

Object.entries(object).forEach(([k, v]) => setValue(result, k, v));

console.log(result);

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