javascript - 如何从多个嵌套对象中计算特定值的对象数量？

Question

var foodwebsites = {
  "bacon": [{
    "url": "stackoverflow.com",
  }],
  "icecream": [{
    "url": "example.com",
  }],
  "cheese": [{
    "url": "example.com",
  }]
}

var baconfoodwebsites = foodwebsites.bacon.filter(function(elem) {
  return elem.url == 'example.com';
}).length;

var icecreamfoodwebsites = foodwebsites.icecream.filter(function(elem) {
  return elem.url == 'example.com';
}).length;

var cheesefoodwebsites = foodwebsites.cheese.filter(function(elem) {
  return elem.url == 'example.com';
}).length;

var allfoodwebsites = baconfoodwebsites + icecreamfoodwebsites + cheesefoodwebsites;

console.log(baconfoodwebsites, icecreamfoodwebsites, cheesefoodwebsites, allfoodwebsites)

我想做完全相同的事情，而不重复所有这些单独的嵌套对象(培根、冰淇淋和奶酪)。

我认为答案是这样的:

var allfoodwebsites = foodwebsites=.filter(function( elem) {
    return elem.url == 'example.com';
}).length;

其他信息:

如果可能的话，我只想使用 jQuery + Pure Javascript。

我想查找所有带有 "url": "example.com"的嵌套对象

Answer 1

使用 Object.values、Array#reduce() 和 Array#flat() 。注意 flat() 在某些环境下可能需要 polyfill

let foodwebsites = {"bacon": [{"url": "stackoverflow.com",}],"icecream": [{"url": "example.com",}],"cheese": [{"url": "example.com",}]};


const  getUrlCount = (url) => {
   return Object.values(foodwebsites)
       .flat()
       .reduce((a, {url:u})=> a + (url === u) , 0)

}

console.log(getUrlCount("example.com"))