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具有递归关系的javascript过滤器

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margherita pizza · 技术社区 · 6 年前

我的用例是这样的。

我有一个称为位置的对象数组。
它包含有关省、区和市的信息。
市属于区,区属于省。
如果 parent_region_id null 意味着它是一个省。
如果 父区域ID 是一个省的ID表示它是一个地区。
否则它就是一个城市。

我的目标是从阵列中筛选出所有城市。我做了一个奇怪的方式,但它工作。

但我想知道什么是最理想的方法来做到这一点? 这是我的工作代码。

const locations = [
{id:1,name:"Western province",parent_region_id:null},
{id:2,name:"Southern province",parent_region_id:null},
{id:3,name:"Central province",parent_region_id:null},
{id:4,name:"Colombo district",parent_region_id:1},
{id:5,name:"Galle district",parent_region_id:2},
{id:6,name:"Kandy district",parent_region_id:3},
{id:7,name:"Maharagama",parent_region_id:4},
{id:8,name:"Nugegoda",parent_region_id:4},
{id:9,name:"Peradeniya",parent_region_id:6},
]

//get the province ids first

const provinces = []

locations.forEach(e=> {
if(!e.parent_region_id){
  provinces.push(e.id)
}
})

// get all the districts then

const districts = []

locations.forEach(e=>{
  if(provinces.includes(e.parent_region_id)){
  districts.push(e.id) 
  }
})


//get cities

const cities = [];


locations.forEach(e=> 
{
  if(!districts.includes(e.id) && !provinces.includes(e.id)){
    console.log(e.name," is a city")
  }
})

2 回复 | 直到 6 年前

kockburn 6 年前

使用 Array#filter 和 Array#find .

const locations=[{id:1,name:"Western province",parent_region_id:null},{id:2,name:"Southern province",parent_region_id:null},{id:3,name:"Central province",parent_region_id:null},{id:4,name:"Colombo district",parent_region_id:1},{id:5,name:"Galle district",parent_region_id:2},{id:6,name:"Kandy district",parent_region_id:3},{id:7,name:"Maharagama",parent_region_id:4},{id:8,name:"Nugegoda",parent_region_id:4},{id:9,name:"Peradeniya",parent_region_id:6},]

const res = locations.filter(({id, parent_region_id},i,a)=>{
  if(!parent_region_id) return false;
  const loc = a.find(l=>l.id === parent_region_id);
  return !loc.parent_region_id ? false : true;
});

console.log(res);

解决方法是先删除省份:

const locations=[{id:1,name:"Western province",parent_region_id:null},{id:2,name:"Southern province",parent_region_id:null},{id:3,name:"Central province",parent_region_id:null},{id:4,name:"Colombo district",parent_region_id:1},{id:5,name:"Galle district",parent_region_id:2},{id:6,name:"Kandy district",parent_region_id:3},{id:7,name:"Maharagama",parent_region_id:4},{id:8,name:"Nugegoda",parent_region_id:4},{id:9,name:"Peradeniya",parent_region_id:6},]

const res = locations
.filter(({parent_region_id})=>parent_region_id!==null)
.filter(({parent_region_id},i,a)=>{
  return a.findIndex(l=>l.id === parent_region_id) > -1;
});

console.log(res);

Jack Bashford 6 年前

如果要使代码更清晰、更紧凑,请尝试使用 filter() 像这样:

 

const locations = [
{id:1,name:"Western province",parent_region_id:null},
{id:2,name:"Southern province",parent_region_id:null},
{id:3,name:"Central province",parent_region_id:null},
{id:4,name:"Colombo district",parent_region_id:1},
{id:5,name:"Galle district",parent_region_id:2},
{id:6,name:"Kandy district",parent_region_id:3},
{id:7,name:"Maharagama",parent_region_id:4},
{id:8,name:"Nugegoda",parent_region_id:4},
{id:9,name:"Peradeniya",parent_region_id:6},
];

const provinces = locations.filter(e => !e.parent_region_id);

const districts = locations.filter(e => provinces.some(p => p.id == e.parent_region_id));

const cities = locations.filter(e => !provinces.includes(e) && !districts.includes(e));

console.log(provinces);
console.log(districts);
console.log(cities);