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

为了完成其他答案,在lodash 4中只忽略undefined和null(而不是像false这样的属性),你可以在_.pickBy中使用谓词:

_。pickBy(obj, v !== null && v !== undefined)

例子如下: Const obj = {a: undefined, b: 123, c: true, d: false, e: null}; const filteredObject = _。pickBy(obj, v => v !== null && v !== undefined); console.log = (obj) => document.write(JSON. log)stringify(filteredObject, null, 2)); console.log (filteredObject); < script src = " https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.js " > < /脚本>

其他回答

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

从对象中删除未定义、空字符串和空字符串

_.omitBy(object, (v) => _.isUndefined(v) || _.isNull(v) || v === '');

如果你想移除所有假值,那么最紧凑的方法是:

对于Lodash 4。X及以后:

_.pickBy({ a: null, b: 1, c: undefined }, _.identity);
>> Object {b: 1}

对于遗留的Lodash 3.x:

_.pick(obj, _.identity);

_.pick({ a: null, b: 1, c: undefined }, _.identity);
>> Object {b: 1}

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

为了完成其他答案,在lodash 4中只忽略undefined和null(而不是像false这样的属性),你可以在_.pickBy中使用谓词:

_。pickBy(obj, v !== null && v !== undefined)

例子如下: Const obj = {a: undefined, b: 123, c: true, d: false, e: null}; const filteredObject = _。pickBy(obj, v => v !== null && v !== undefined); console.log = (obj) => document.write(JSON. log)stringify(filteredObject, null, 2)); console.log (filteredObject); < script src = " https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.js " > < /脚本>