如何在es6中过滤嵌套对象并返回父对象

Question

I have an object that contains nested objects. i would like to filter through them and return the key if present.

for example:

var meals = {
    food_meals: [
      {meal_id: 15749, address_required: false, button_text: "choose", can_choose_meal: true},
      {meal_id: 15750, address_required: false, button_text: "choose", can_choose_meal: true}
    ],
    wine_meals: [
      {meal_id: 11651, address_required: false, button_text: "choose", can_choose_meal: true},
      {meal_id: 4424, address_required: false, button_text: "choose", can_choose_meal: true}
    ],
    kids_meals: [
      {meal_id: 15763, address_required: false, button_text: "choose", can_choose_meal: true},
      {meal_id: 15764, address_required: false, button_text: "choose", can_choose_meal: true},
      {meal_id: 15765, address_required: false, button_text: "choose", can_choose_meal: true}
    ]
}

If i had a meal with meal_id of 15764 then I would want the key of that value returned (in this case kids_meals )

i can filter the meal from the nested object by doing

meals.kids_meals.filter(meal => meal.meal_id == this.props.selection.meal_id)

where this.props.selection.meal_id is 15764

my desired output is 'kids_meals' in this case, but i can't seem to reach it

Answer 1

You can iterate over the Object.entries of the meals to get an array of key-value pairs, and use .find on that array to get the key-value pair whose value is an array that contains the matching meal_id :

 const meals = { food_meals: [ {meal_id: 15749, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15750, address_required: false, button_text: "choose", can_choose_meal: true} ], wine_meals: [ {meal_id: 11651, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 4424, address_required: false, button_text: "choose", can_choose_meal: true} ], kids_meals: [ {meal_id: 15763, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15764, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15765, address_required: false, button_text: "choose", can_choose_meal: true} ] }; const idToFind = 15764; const foundEntry = Object.entries(meals) .find( ([, arr]) => arr.some( ({ meal_id }) => meal_id === idToFind ) ); if (foundEntry) { console.log(foundEntry[0]); } 

Answer 2

The function filter is for creating a new array with objects which meet a specific condition. In this case, you need to use the function find to extract an object which meets a condition.

This alternative uses the function find along with the function some to find at least one object with a specific meal_id .

 let meals = { food_meals: [ {meal_id: 15749, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15750, address_required: false, button_text: "choose", can_choose_meal: true} ], wine_meals: [ {meal_id: 11651, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 4424, address_required: false, button_text: "choose", can_choose_meal: true} ], kids_meals: [ {meal_id: 15763, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15764, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15765, address_required: false, button_text: "choose", can_choose_meal: true} ]}, target = 15764, found = Object.keys(meals).find(k => meals[k].some(({meal_id}) => meal_id === target)); console.log(found); 

Answer 3

Using Array#filter , Array#findIndex and Object.keys

 const data={food_meals:[{meal_id:15749,address_required:!1,button_text:"choose",can_choose_meal:!0},{meal_id:15750,address_required:!1,button_text:"choose",can_choose_meal:!0}],wine_meals:[{meal_id:11651,address_required:!1,button_text:"choose",can_choose_meal:!0},{meal_id:4424,address_required:!1,button_text:"choose",can_choose_meal:!0}],kids_meals:[{meal_id:15763,address_required:!1,button_text:"choose",can_choose_meal:!0},{meal_id:15764,address_required:!1,button_text:"choose",can_choose_meal:!0},{meal_id:15765,address_required:!1,button_text:"choose",can_choose_meal:!0}]} const id = 15764; const res = Object.keys(data).filter(key=>{ return data[key].findIndex(({meal_id})=>meal_id===id) > -1 }).join(""); console.log(res); 

Answer 4

I realize this is relatively close to @CertainPerformance 's answer, and I don't mean to say that his is invalid at all, however, you might find that this is a bit easier to read, and secondly if there's ever a meal that is available in more than one category( a use case that you may not have thought of ), it'll return an array.

Object.keys(meals).filter(type => meals[type].some(item => item.meal_id === idToFind));

 var meals = { food_meals: [ {meal_id: 15749, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15750, address_required: false, button_text: "choose", can_choose_meal: true} ], wine_meals: [ {meal_id: 11651, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 4424, address_required: false, button_text: "choose", can_choose_meal: true} ], kids_meals: [ {meal_id: 15763, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15764, address_required: false, button_text: "choose", can_choose_meal: true}, {meal_id: 15765, address_required: false, button_text: "choose", can_choose_meal: true} ] } let r = Object.keys(meals).filter(type => meals[type].some(item => item.meal_id === 15749)); console.log(r);