如何删除JavaScript对象中未定义或空的所有属性?

(这个问题与数组的问题类似)


当前回答

这是另一种选择

打字稿:

function objectDefined <T>(obj: T): T {
  const acc: Partial<T> = {};
  for (const key in obj) {
    if (obj[key] !== undefined) acc[key] = obj[key];
  }
  return acc as T;
}

Javascript:

function objectDefined(obj) {
  const acc = {};
  for (const key in obj) {
    if (obj[key] !== undefined) acc[key] = obj[key];
  }
  return acc;
}

其他回答

ES6 arrow function and ternary operator:
Object.entries(obj).reduce((acc, entry) => {
   const [key, value] = entry
  if (value !== undefined) acc[key] = value;
  return acc;
}, {})
    const obj = {test:undefined, test1:1 ,test12:0, test123:false};
    const newObj = Object.entries(obj).reduce((acc, entry) => {
       const [key, value] = entry
      if (value !== undefined) acc[key] = value;
      return acc;
    }, {})
    console.log(newObj)

清洁物体到位

// General cleanObj function
const cleanObj = (valsToRemoveArr, obj) => {
   Object.keys(obj).forEach( (key) =>
      if (valsToRemoveArr.includes(obj[key])){
         delete obj[key]
      }
   })
}

cleanObj([undefined, null], obj)

纯函数

const getObjWithoutVals = (dontReturnValsArr, obj) => {
    const cleanObj = {}
    Object.entries(obj).forEach( ([key, val]) => {
        if(!dontReturnValsArr.includes(val)){
            cleanObj[key]= val
        } 
    })
    return cleanObj
}

//To get a new object without `null` or `undefined` run: 
const nonEmptyObj = getObjWithoutVals([undefined, null], obj)

TypeScript的泛型函数

function cleanProps(object:Record<string, string>):Record<string, string> {
  let cleanObj = {};

  Object.keys(object).forEach((key) => {
    const property = object[key];
    cleanObj = property ? { ...cleanObj, [key]: property } : cleanObj;
  });

  return cleanObj;
}

export default cleanProps;

现在假设你有一个像下面这样的对象

interface Filters{
 searchString: string;
 location: string;
 sector: string
}

const filters:Filters = {
  searchString: 'cute cats',
  location: '',
  sector: 'education',
};

您可以按照如下方式使用该函数

const result = cleanProps(filters as Record<keyof Filters, string>);
console.log(result); // outputs: { searchString: 'cute cats', sector: 'education' }

您可能正在寻找delete关键字。

var obj = { };
obj.theProperty = 1;
delete obj.theProperty;

如果你正在使用lodash或underscore.js,这里有一个简单的解决方案:

var obj = {name: 'John', age: null};

var compacted = _.pickBy(obj);

这将只适用于lodash 4,预lodash 4或下划线。js,使用_。选择(obj _.identity);