如何展平 Typescript 中的嵌套对象？

Question

我正在尝试将以下 object 'raw' 展平为 object 'flat'

raw = [    {
      "id":"123",
      "Date":"12/12/2020",
      "Type":{
         "id":"456",
         "desc":"test1"
      }    },    {
      "id":"124",
      "Date":"12/12/2020",
      "Type":{
         "id":"456",
         "desc":"test2"
      }    } ]

flat =[   {
      "id":"123",
      "Date":"12/12/2020",
      "desc":"test1"
      },    
      {
      "id":"124",
      "Date":"12/12/2020",
      "desc":"test2"
      }]

我尝试了以下操作：

   let flatData:any = []
    const flattenObject = (obj:any) => {
      const flattened:any = {}
    
      Object.keys(obj).forEach((key) => {
        if (typeof obj[key] === 'object' && obj[key] !== null) {
          Object.assign(flattened, flattenObject(obj[key]))
        } else {
          flattened[key] = obj[key]
        }
      })
      flatData.push(flattened)
      console.log(flattened)
      return flattened
    }

结果我从上面的代码片段中得到。

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Answer 1

您可以在 TypeScript 中使用object 解构。

 const raw = [{ "id": "123", "Date": "12/12/2020", "Type": { "id": "456", "desc": "test1" } }, { "id": "124", "Date": "12/12/2020", "Type": { "id": "456", "desc": "test2" } }]; const mapped = raw.map(({ id, Date, Type: { desc } }) => ({ id, Date, desc })); console.log(mapped);

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Answer 2

您用于展平 object 的 function 是正确的，但是Type属性中的嵌套id属性与顶级id属性具有相同的属性名称。 当您展平 object 时，该顶级id值将被嵌套的id值覆盖。

解决方案：

1. 如果您可以控制数据，则可以将嵌套的id属性重命名为其他名称。
2. 在flattenObject function 中，您可以使用父属性名称作为嵌套属性的前缀。 IE

const flattenObject = (obj:any, prefix = '') => {
      const flattened:any = {}
    
      Object.keys(obj).forEach((key) => {
        if (typeof obj[key] === 'object' && obj[key] !== null) {
          Object.assign(flattened, flattenObject(obj[key], prefix))
        } else {
          flattened[prefix + key] = obj[key]
        }
      })
      flatData.push(flattened)
      return flattened

Answer 3

如果您只有一个深度级别并且想要展平对象，无论属性名称是什么，这是一个有效的解决方案：

 const raw = [ { id: "123", Date: "12/12/2020", Type: { id: "456", desc: "test1", }, }, { id: "124", Date: "12/12/2020", Type: { id: "456", desc: "test2", }, }, ]; const flatten = (obj) => Object.assign( {}, Object.fromEntries( Object.values(obj).filter((x) => typeof x === "object").map((x) => Object.entries(x)).flat(1) ), Object.fromEntries( Object.entries(obj).filter(([, x]) => typeof x;== "object") ) ). const compute = (data) => data;map(flatten). console;log(compute(raw));