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

最简单的Lodash解决方案返回一个过滤掉空值和未定义值的对象。

_.omitBy(obj, _.isNil)

其他回答

下面是一个综合递归函数(最初基于@chickens的函数),它将:

递归删除你告诉它的默认值=[undefined, null, ", NaN] 正确处理常规对象、数组和Date对象

const cleanEmpty = function(obj, defaults = [undefined, null, NaN, '']) {
  if (!defaults.length) return obj
  if (defaults.includes(obj)) return

  if (Array.isArray(obj))
    return obj
      .map(v => v && typeof v === 'object' ? cleanEmpty(v, defaults) : v)
      .filter(v => !defaults.includes(v))

  return Object.entries(obj).length 
    ? Object.entries(obj)
        .map(([k, v]) => ([k, v && typeof v === 'object' ? cleanEmpty(v, defaults) : v]))
        .reduce((a, [k, v]) => (defaults.includes(v) ? a : { ...a, [k]: v}), {}) 
    : obj
}

用法:

// based off the recursive cleanEmpty function by @chickens. // This one can also handle Date objects correctly // and has a defaults list for values you want stripped. const cleanEmpty = function(obj, defaults = [undefined, null, NaN, '']) { if (!defaults.length) return obj if (defaults.includes(obj)) return if (Array.isArray(obj)) return obj .map(v => v && typeof v === 'object' ? cleanEmpty(v, defaults) : v) .filter(v => !defaults.includes(v)) return Object.entries(obj).length ? Object.entries(obj) .map(([k, v]) => ([k, v && typeof v === 'object' ? cleanEmpty(v, defaults) : v])) .reduce((a, [k, v]) => (defaults.includes(v) ? a : { ...a, [k]: v}), {}) : obj } // testing console.log('testing: undefined \n', cleanEmpty(undefined)) console.log('testing: null \n',cleanEmpty(null)) console.log('testing: NaN \n',cleanEmpty(NaN)) console.log('testing: empty string \n',cleanEmpty('')) console.log('testing: empty array \n',cleanEmpty([])) console.log('testing: date object \n',cleanEmpty(new Date(1589339052 * 1000))) console.log('testing: nested empty arr \n',cleanEmpty({ 1: { 2 :null, 3: [] }})) console.log('testing: comprehensive obj \n', cleanEmpty({ a: 5, b: 0, c: undefined, d: { e: null, f: [{ a: undefined, b: new Date(), c: '' }] }, g: NaN, h: null })) console.log('testing: different defaults \n', cleanEmpty({ a: 5, b: 0, c: undefined, d: { e: null, f: [{ a: undefined, b: '', c: new Date() }] }, g: [0, 1, 2, 3, 4], h: '', }, [undefined, null]))

JSON。Stringify删除未定义的键。

removeUndefined = function(json){
  return JSON.parse(JSON.stringify(json))
}

我在我的项目中有同样的场景,并使用以下方法实现。

它适用于所有数据类型,上面提到的一些数据类型不适用于日期和空数组。

removeEmptyKeysFromObject.js

removeEmptyKeysFromObject(obj) { Object.keys(obj).forEach(key => { if (Object.prototype.toString.call(obj[key]) === '[object Date]' && (obj[key].toString().length === 0 || obj[key].toString() === 'Invalid Date')) { delete obj[key]; } else if (obj[key] && typeof obj[key] === 'object') { this.removeEmptyKeysFromObject(obj[key]); } else if (obj[key] == null || obj[key] === '') { delete obj[key]; } if (obj[key] && typeof obj[key] === 'object' && Object.keys(obj[key]).length === 0 && Object.prototype.toString.call(obj[key]) !== '[object Date]') { delete obj[key]; } }); return obj; }

将任何对象传递给该函数

30多个答案,但我没有看到这个简短的ES6一行程序,由于Object.assign()是一个无声地忽略任何非对象(如false)的变参数函数,因此利用了扩散操作符。

Object.assign({}, ...Object.entries(obj).map(([k,v]) => v != null && {[k]: v]))

你可以剪短一点!条件

var r = {a: null, b: undefined, c:1};
for(var k in r)
   if(!r[k]) delete r[k];

使用时请记住:as @分色announcement in comments:如果值为空字符串、false或0,这也会删除属性