javascript - 引用父对象展平嵌套对象

Question

我想压平下面的

let o = {
  name: "John",
  school: {
    name: "Phillps",
  }
};

至:

{
  name: "John",
  schoolName: "Phillps"
}

我的代码如下所示

f= Object.assign({}, ..._flatten(o));

function _flatten(o) {
  return [].concat(
    ...Object.keys(o).map(k =>
      typeof o[k] === "object" ? _flatten(o[k]) : { [k]: o[k] }
    )
  );
}

这会产生

{  
 name: "Phillps"
}

如您所见，它无法处理嵌套对象中属性名称的冲突。 IE。学生的名字被学校的名字覆盖。是否可以修改代码，使深层属性以其父对象名称为前缀？

Answer 1

您可以使用Object.entries , Array.prototype.flatMap和 Object.fromEntries

const upperFirst = (str = "") =>
  str[0] .toUpperCase () + str.substr(1)

const camelCase = ([ first = "", ...rest ]) =>
  first + rest .map (upperFirst) .join ('')

const append = (xs, x) =>
  xs .concat ([ x ])
  
const flatten = (o = {}) =>
{ const loop = (o, path) =>
    Object (o) === o
      ?  Object .entries (o) .flatMap
           ( ([ k, v ]) =>
               loop
                 ( v
                 , append (path, k)
                 )
           )
      : [ [ camelCase (path), o ] ]
  return Object .fromEntries (loop (o, []))
}

console.log
  ( flatten
      ( { name: "John"
        , school:
            { name: "Phillips"
            , district: { zone: 1 }
            }
        }
      )
  )
  
// { "name": "John"
// , "schoolName": "Phillips"
// , "schoolDistrictZone": 1
// }

flatMap 急切地评估输入并在 flatten 返回之前创建一些中间值。因为 Object.fromEntries 接受任何可迭代对象，所以我们最好使用生成器编写 loop

const flatten = (o = {}) =>
{ const loop = function* (o, path)
  { if (Object (o) === o)
      for (const [ k, v ] of Object .entries (o))
        yield* loop
          ( v
          , append (path, k)
          )
    else
      yield [ camelCase (path), o ]
  }
  return Object .fromEntries (loop (o, []))
}

重新运行程序，您将看到完全相同的输出。另外值得一提的是两个程序之间的结构相似性。