Question

我想做点像......

_.extend(original, fresh);

但只能复制非空的属性，所以......

var original = {a:1, b:2, c:3}
var fresh = {a:5, b:null}

var mixed = _.extend(original, fresh)
// currently mixed is {a:5,b:null,c:3}
// I want it to be {a:5,b:2,c:3}

更新

我取得了一些进展，但它不适用于嵌套对象......

http://jsfiddle.net/billymoon/2GwMx/1/

var original = {a:1, b:2, c:3, d:{g:1}}
var fresh = {a:5, b:null,c:"", d:{e:4,f:5,g:null}}

var mixed = _.extend( original, _.reduce(fresh,function(a,v,k){ 
     if(v){ a[k]=v; } 
     return a; 
}, {}) )

// currently mixed is {"a":5,"b":2,"c":3,"d":{"e":4,"f":5,"g":null}}
// I want it to be {"a":5,"b":2,"c":3,"d":{"e":4,"f":5,g:1}}
alert(JSON.stringify(mixed))

Answer 1

一点带有下划线的javascript：

_.mixin({
  extendNonNull: function(obj) {
    _.each(_.rest(arguments,1), function(source) {
      if (source) {
        for (var prop in source) {
          if( _.isObject(source[prop]) && _.isObject(obj[prop])){
             obj[prop] = _.extendNonNull(obj[prop], source[prop]);
          }
          else if( !_.isNull(source[prop])){
             obj[prop] = source[prop];
          }
        }
      }
    });
    return obj;
  }
});

var original = {a:1, b:2, c:3, d:{g:1}};
var fresh = {a:5, b:null,c:"", d:{e:4,f:5,g:null}};

var mixed = _.extendNonNull(original, fresh);

Answer 2

如果你使用lodash或下划线，你可以这样做：

import _ from 'lodash';
let result = _.merge({},target, source);
let clonedNonNull = _.omit(result,_.isNull);

Answer 3

使用Lodash，您可以将过滤器功能作为 assignWith 功能的最后一个参数传递。
然后，您可以过滤 null 值或更具体的案例，并同时合并两个以上的对象。

let obj1 = {a: 'foo', b: null}
let obj2 = {a: null, b: 'bar'}

// acc is the accumulator value
function filter(acc, newValue) {
  let isNull = _.isNull(newValue);
  return isNull ? acc : newValue;
}

let mixin = _.assignWith({}, obj1, obj2, filter); // {a: 'foo', b: 'bar'}

Underscore JS使用来自新对象的非null值扩展对象

更新

3 个答案: