根据包含来自另一个数组的所有值的数组 属性 过滤对象数组

Filter an array of objects based on array property containing all values from another array

我有一个对象数组:

const breeds=[{name: 'Golden', temperament: ['friendly', 'kind', 'smart']},{name: 'Husky', temperament: ['alert', 'loyal', 'gentle']},{name: 'Yorkshire Terrier', temperament: ['bold', 'independent', 'kind']}]

我想按选择对它们进行排序 "temperament." 假设用户同时选择了 "kind" 和 "friendly",它应该只 return "Golden" .

我正在使用 javascript 和下划线,这是我迄今为止尝试过的方法:

 //selected is an array of selected temperaments
 //breeds is the array of objects
function filterTemperaments(selected, breeds) {
  return _.filter(breeds, function (breed) {
    if (!breed.temperament) breed.temperament = "";
    const breedList = breed.temperament;
    return breedList.includes(...selected);
  }, selected);
}

这似乎只是 returning 符合所选数组中第一个气质的品种。例如,如果选择的是 ['kind', 'loyal'] 并且品种是 {name:'Golden', temperament: ['kind', 'smelly']},尽管不符合 "Loyal" 气质

,Golden 仍然会返回 true

这里有更好的解决方案吗?提前致谢!!

可以使用filter to return only those breeds which have every气质选择

const breeds=[{name: 'Golden', temperament: ['friendly', 'kind', 'smart']},{name: 'Husky', temperament: ['alert', 'loyal', 'gentle']},{name: 'Yorkshire Terrier', temperament: ['bold', 'independent', 'kind']}],
    selected = ['kind', 'friendly']

const filtered = breeds.filter(b => selected.every(s => b.temperament.includes(s)))

console.log(filtered)

使用 filterevery 可以很容易地使用它来了解您是否有任何匹配以及 return 匹配的值。

const breeds = [{
  name: 'Golden',
  temperament: ['friendly', 'kind', 'smart']
}, {
  name: 'Husky',
  temperament: ['alert', 'loyal', 'gentle']
}, {
  name: 'Yorkshire Terrier',
  temperament: ['bold', 'independent', 'kind']
}]

//selected is an array of selected temperaments
//breeds is the array of objects
function filterTemperaments(selected, breeds) {
  return breeds.filter(({ temperament }) => { //here I'm desctructuring the object only to get the temperaments.
    return selected.every(selection => temperament.indexOf(selection) !== -1)
  })

}

const result = filterTemperaments(['kind', 'friendly'], breeds);
console.log(result)