我有一个Javascript对象像:

var my_object = { a:undefined, b:2, c:4, d:undefined };

如何删除所有未定义的属性?False属性应该保留。


当前回答

我喜欢用_。pickBy,因为你可以完全控制你要删除的东西:

var person = {"name":"bill","age":21,"sex":undefined,"height":null};

var cleanPerson = _.pickBy(person, function(value, key) {
  return !(value === undefined || value === null);
});

来源:https://www.codegrepper.com/?search_term=lodash +删除+未定义值+ + +对象

其他回答

如果您不想删除假值。这里有一个例子:

obj = {
  "a": null,
  "c": undefined,
  "d": "a",
  "e": false,
  "f": true
}
_.pickBy(obj, x => x === false || x)
> {
    "d": "a",
    "e": false,
    "f": true
  }

考虑到undefined == null,我们可以这样写:

let collection = {
  a: undefined,
  b: 2,
  c: 4,
  d: null,
}

console.log(_.omit(collection, it => it == null))
// -> { b: 2, c: 4 }

JSBin例子

用于深嵌套的对象和数组。并从字符串和NaN中排除空值

function isBlank(value) {
  return _.isEmpty(value) && !_.isNumber(value) || _.isNaN(value);
}
var removeObjectsWithNull = (obj) => {
  return _(obj).pickBy(_.isObject)
    .mapValues(removeObjectsWithNull)
    .assign(_.omitBy(obj, _.isObject))
    .assign(_.omitBy(obj, _.isArray))
    .omitBy(_.isNil).omitBy(isBlank)
    .value();
}
var obj = {
  teste: undefined,
  nullV: null,
  x: 10,
  name: 'Maria Sophia Moura',
  a: null,
  b: '',
  c: {
    a: [{
      n: 'Gleidson',
      i: 248
    }, {
      t: 'Marta'
    }],
    g: 'Teste',
    eager: {
      p: 'Palavra'
    }
  }
}
removeObjectsWithNull(obj)

结果:

{
   "c": {
      "a": [
         {
            "n": "Gleidson",
            "i": 248
         },
         {
            "t": "Marta"
         }
      ],
      "g": "Teste",
      "eager": {
         "p": "Palavra"
      }
   },
   "x": 10,
   "name": "Maria Sophia Moura"
}

默认情况下,pickBy使用identity:

_.pickBy({ a: null, b: 1, c: undefined, d: false });

我也会使用下划线并处理空字符串:

Var my_object = {a:undefined, b:2, c:4, d:undefined, k: null, p: false, s: ", z: 0}; Var结果=_。省略(my_object, function(value) { return _.isUndefined(value) || _.isNull(value) || value === "; }); console.log(结果);//对象{b: 2, c: 4, p: false, z: 0}

JSBIN.