javascript - 将深层对象映射到新对象的更好方法

Question

此代码用于将 JSON 转换为一个对象，其中每个名称对象都变成其值的键，或者如果它有自己的元素对象，则将其分解并对其内容执行相同的操作。

是否有更好的方法来做到这一点，同时允许 JSON 模式具有更大的可扩展性？
有没有一种方法可以将它全部归结为一个更简单的函数，我可以传递第一个元素并将其转换为架构达到的任何深度？

const fs = require('fs');
{
    let scheme = JSON.parse('{"$schema":{"root":{"name":"THINGY","dtd":{"name":"DOCTYPE","value":"something.dtd","commentBefore":["?xml version='1.0'?","Version NULL"]},"ele":{"name":"REPORT","ele":[{"name":"SEGMENT0","ele":[{"name":"NUMBER1","value":""},{"name":"NUMBER2","value":""}]},{"name":"SEGMENT1","ele":[{"name":"RECORD1","ele":[{"name":"NUMBER1","value":""},{"name":"NUMBER2","value":""}]}]},{"name":"SEGMENT2","ele":[]},{"name":"SEGMENT3","ele":[]},{"name":"SEGMENT4","ele":[]},{"name":"SEGMENT5","ele":[]}]}}}}').$schema.root;
    let depth = 0;

    var compiled = {
        [scheme.ele.name]: scheme.ele.ele.map(function(i) {
            if (typeof i.ele != 'undefined') {
                return {
                    [i.name]: i.ele.map(function(k) {
                        if (typeof k.ele != 'undefined') {
                            return {
                                [k.name]: k.ele.map(function(p) {
                                    if (typeof p.ele != 'undefined') {
                                        return {
                                            [p.name]: p.ele
                                        };
                                    } else {
                                        return {
                                            [p.name]: p.value
                                        };
                                    }
                                })
                            };
                        } else {
                            return {
                                [k.name]: k.value
                            };
                        }
                    })
                };
            } else {
                return {
                    [i.name]: i.value
                };
            }
        })
    };
}

console.log(JSON.stringify(compiled, 0, 2));

我应该补充一点，这旨在最终也应用验证并在到达字符串对象时获取真实数据。

输出看起来像这样:

{
    "REPORT": [
    {
        "SEGMENT0": [
        {
            "NUMBER1": ""
        },
        {
            "NUMBER2": ""
        }
        ]
    },
    {
        "SEGMENT1": [
        {
            "RECORD1": [
            {
                "NUMBER1": ""
            },
            {
                "NUMBER2": ""
            }
            ]
        }
        ]
    },
    {
        "SEGMENT2": []
    },
    {
        "SEGMENT3": []
    },
    {
        "SEGMENT4": []
    },
    {
        "SEGMENT5": []
    }
    ]
}

Answer 1

您可以解构对象，获取name、ele 和value 并返回一个带有name 的新对象作为键和通过映射 ele 或 value 对象的数组。

const
    getData = ({ name, ele, value }) => ({
        [name]: Array.isArray(ele)
            ? ele.map(getData)
            : value
    });

var scheme = JSON.parse('{"$schema":{"root":{"name":"THINGY","dtd":{"name":"DOCTYPE","value":"something.dtd","commentBefore":["?xml version=\'1.0\'?","Version NULL"]},"ele":{"name":"REPORT","ele":[{"name":"SEGMENT0","ele":[{"name":"NUMBER1","value":""},{"name":"NUMBER2","value":""}]},{"name":"SEGMENT1","ele":[{"name":"RECORD1","ele":[{"name":"NUMBER1","value":""},{"name":"NUMBER2","value":""}]}]},{"name":"SEGMENT2","ele":[]},{"name":"SEGMENT3","ele":[]},{"name":"SEGMENT4","ele":[]},{"name":"SEGMENT5","ele":[]}]}}}}').$schema.root,
    result = getData(scheme.ele);

console.log(result);

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