ES6+的最短一行

过滤所有错误值(""，0,false, null, undefined)

Object.entries(obj).reduce((a,[k,v]) => (v ? (a[k]=v, a) : a), {})

过滤空值和未定义值:

Object.entries(obj).reduce((a,[k,v]) => (v == null ? a : (a[k]=v, a)), {})

仅过滤null

Object.entries(obj).reduce((a,[k,v]) => (v === null ? a : (a[k]=v, a)), {})

过滤器仅未定义

Object.entries(obj).reduce((a,[k,v]) => (v === undefined ? a : (a[k]=v, a)), {})

递归解决方案:过滤null和undefined

对象:

const cleanEmpty = obj => Object.entries(obj)
        .map(([k,v])=>[k,v && typeof v === "object" ? cleanEmpty(v) : v])
        .reduce((a,[k,v]) => (v == null ? a : (a[k]=v, a)), {});

对于对象和数组:

const cleanEmpty = obj => {
  if (Array.isArray(obj)) { 
    return obj
        .map(v => (v && typeof v === 'object') ? cleanEmpty(v) : v)
        .filter(v => !(v == null)); 
  } else { 
    return Object.entries(obj)
        .map(([k, v]) => [k, v && typeof v === 'object' ? cleanEmpty(v) : v])
        .reduce((a, [k, v]) => (v == null ? a : (a[k]=v, a)), {});
  } 
}

JSON。Stringify删除未定义的键。

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

我们可以使用JSON。stringify和JSON。解析以从对象中删除空白属性。

jsObject = JSON.parse(JSON.stringify(jsObject), (key, value) => {
               if (value == null || value == '' || value == [] || value == {})
                   return undefined;
               return value;
           });

Lodash:

_.omitBy({a: 1, b: null}, (v) => !v)

如果有人需要欧文(和埃里克)答案的递归版本，这里是:

/**
 * Delete all null (or undefined) properties from an object.
 * Set 'recurse' to true if you also want to delete properties in nested objects.
 */
function delete_null_properties(test, recurse) {
    for (var i in test) {
        if (test[i] === null) {
            delete test[i];
        } else if (recurse && typeof test[i] === 'object') {
            delete_null_properties(test[i], recurse);
        }
    }
}

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]))