如果他的属性匹配数组中的所有项目（例如食谱的成分），则返回 object

Question

I'm learning js and I was trying to filter recipes based on the ingredients I have inside an array but the check done with Array.every() seems to not work, what am I doing wrong? Here's the code, thanks all!

const plates = [
  {
    name: "carbonara",
    ingredients: ["eggs", "milk", "pepper", "pasta", "cheese"],
    time: 30,
    difficulty: "easy",
  },
  {
    name: "oil_pepper",
    ingredients: ["oil", "pepper", "pasta", "garlic"],
    time: 20,
    difficulty: "medium",
  },
  {
    name: "tomato",
    ingredients: ["pasta", "tomato", "garlic", "onion"],
    time: "30",
    difficulty: "hard",
  },
];

const fridge = ["oil", "pepper", "pasta", "garlic"];


let filteredPlates = [];

const filter = () => {
  plates.forEach((element) =>
    fridge.forEach((element2) => {
      if (
        element.ingredients.every((element3) =>
          element3.includes(element2.toString())
        )
      ) {
        filteredPlates.push(element);
        return filteredPlates;
      }
    })
  );
};

filter();

console.log(filteredPlates);

Answer 1

You are over-complicating the problem. You just need to check for each plate, whether every ingredient in that plate's recipe is included in the fridge. You can use Array.filter for this purpose:

 const plates = [ { name: "carbonara", ingredients: ["eggs", "milk", "pepper", "pasta", "cheese"], time: 30, difficulty: "easy", }, { name: "oil_pepper", ingredients: ["oil", "pepper", "pasta", "garlic"], time: 20, difficulty: "medium", }, { name: "tomato", ingredients: ["pasta", "tomato", "garlic", "onion"], time: "30", difficulty: "hard", }, ]; const fridge = ["oil", "pepper", "pasta", "garlic"]; const filteredPlates = plates.filter(plt => plt.ingredients.every(ing => fridge.includes(ing))) console.log(filteredPlates)