将深层对象映射到新对象的更好方法

Question

这段代码用于将JSON转换为一个对象，其中每个名称对象都将变成其值的键，或者如果它具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。

有没有更好的方法来做到这一点，同时还能使JSON模式更具可扩展性？
有没有办法将所有内容简化为一个更简单的函数，即可以传递第一个元素并将其转换为架构所经过的任何深度？

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，这旨在最终还应用验证并在到达字符串对象时获取实际数据。

输出看起来像这样：

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

Answer 1

你可以解构的对象，拿name，ele并value与返回一个新的对象name为重点，要么阵列通过映射的对象ele或value。

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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