Javascript / React Native - 如何在两个不同的对象数组之间进行交叉检查？答案

【问题标题】：Javascript / React Native - How can I make a cross check between two different array of objects?Javascript / React Native - 如何在两个不同的对象数组之间进行交叉检查？
【发布时间】：2021-01-25 02:49:34
【问题描述】：

我有一个应用程序，它将创建一个食谱数据库，用户将在其中输入他拥有的成分，并且该应用程序应该返回用户输入的这种组合可用的食谱。

一旦我拥有包含许多键/值信息的配方对象：

 const [recipes, setRecipes] = useState([
    { id: 1, easyOfPrepare: 'facil', author: 'www.', category: 'Salgada', type: 'Prato principal', name: 'pão de forma', key: '1', ingredients: ['sal', 'ovo', 'mantega', 'maionese', 'farinha'] },
    { id: 2, easyOfPrepare: 'facil', author: 'www.', category: 'Salgada', type: 'Entrada', name: 'macarrao ao molho branco', key: '2', ingredients: ['macarrao', 'oleo', 'creme de leite'] },
    { id: 3, easyOfPrepare: 'facil', author: 'www.', category: 'Salgada', type: 'Salada', name: 'Arroz marroquino', key: '3', ingredients: ['A', 'B', 'C'] },
    { id: 4, easyOfPrepare: 'facil', author: 'www.', category: 'Doce', type: 'Sobremesa', name: 'Bolo', key: '4', ingredients: ['farinha', 'açucar', 'etc'] },
    { id: 5, easyOfPrepare: 'facil', author: 'www.', category: 'Doce', type: 'Bolo', name: 'doce de leite', key: '5', ingredients: ['AA', 'BB', 'CC'] },
  ])

用户输入保存在另一个对象中

const [ingredients, setIngredients] = useState(['farinha', 'ovos', 'leite'])

const addIngredients = e => {
  e.preventDefault();
  setIngredients(prevIngredient => [...prevIngredient, name]);
}

一旦用户通知了预期的成分，就会调用这个添加成分。

一旦用户输入了以下组合：“sal”、“ovo”、“mantega”、“maionese”、“farinha”，我如何根据食谱过滤这些成分以定义唯一可能的成分组合是第一个？

谢谢！

【问题讨论】：

标签： javascript node.js reactjs react-native

【解决方案1】：

利用数组的方法：

const recipes = [
  { id: 1, easyOfPrepare: 'facil', author: 'www.', category: 'Salgada', type: 'Prato principal', name: 'pão de forma', key: '1', ingredients: ['sal', 'ovo', 'mantega', 'maionese', 'farinha'] },
  { id: 2, easyOfPrepare: 'facil', author: 'www.', category: 'Salgada', type: 'Entrada', name: 'macarrao ao molho branco', key: '2', ingredients: ['macarrao', 'oleo', 'creme de leite'] },
  { id: 3, easyOfPrepare: 'facil', author: 'www.', category: 'Salgada', type: 'Salada', name: 'Arroz marroquino', key: '3', ingredients: ['A', 'B', 'C'] },
  { id: 4, easyOfPrepare: 'facil', author: 'www.', category: 'Doce', type: 'Sobremesa', name: 'Bolo', key: '4', ingredients: ['farinha', 'açucar', 'etc'] },
  { id: 5, easyOfPrepare: 'facil', author: 'www.', category: 'Doce', type: 'Bolo', name: 'doce de leite', key: '5', ingredients: ['AA', 'BB', 'CC'] },
];
const userInput = ['sal', 'ovo', 'mantega', 'maionese', 'farinha'];

const result = recipes.filter(recipe => userInput.every(i => recipe.ingredients.includes(i)));


console.log(result.length); // 1

【讨论】：

【解决方案2】：

您可以使用filter 和includes 方法来实现：

const matchedRecipes = recipes.filter(recipe => {
  for (ingredient in ingredients) {
    if (!recipe.ingredients.includes(ingredient)) {
      return false
    }
  }
  return true
})

【讨论】：

【解决方案3】：

您可以根据recipes过滤掉ingredients

const filteredRecipes = recipes.filter(recipe => {
   return ingredients.every(ingredient => recipe.ingredients.includes(ingredient))
})

【讨论】：