javascript - 根据多个搜索词筛选对象数组

Question

假设我有一个搜索词，const searchTerms = ['sarasota', 'fl']

和位置列表，

const selectedRestaurants = [{
  city: "Orlando",
  name: "Dons Food Place",
  state: "FL",
  id: 1
},
{
  city: "Sarasota",
  name: "Rons World",
  state: "FL",
  id: 2
}]

如果我搜索“Sarasota FL”，我只想返回一个 Sarasota FL 位置。但是，如果我基于多个参数(城市和州)进行搜索，则搜索会返回两个位置，因为它与我第二次迭代期间的州相匹配。编辑:当我只搜索“Sarasota”时，我也需要这个来工作

这是我的代码:

filterBySearchTerm(searchTerms) {
   let searchedRestaurants = [];

      for (var j = 0; j < selectedRestaurants.length; j++) {
        searchTerms.forEach(term => {
          if (term.toLowerCase() === this.state.selectedRestaurants[j].city.toLowerCase()) {
            searchedRestaurants.push(this.state.selectedRestaurants[j])
          }
          if (term.toLowerCase() === this.state.selectedRestaurants[j].state.toLowerCase()) {
            searchedRestaurants.push(this.state.selectedRestaurants[j])
          }
          if (term.toLowerCase() === this.state.selectedRestaurants[j].name.toLowerCase()) {
            searchedRestaurants.push(this.state.selectedRestaurants[j])
          }
        })
      }

      //I use Set to remove duplicates ^^
      searchedRestaurants = [...new Set(searchedRestaurants)]
      
      return searchedRestaurants;
  }

我现在已经被困在这个问题上了……有人能帮忙吗？可能没有什么不同，但我在 ReactJS 中。

Answer 1

const restaurants = [
  {
    city: 'Orlando',
    name: 'Dons Food Place',
    state: 'FL',
    id: 1,
  },
  {
    city: 'Sarasota',
    name: 'Rons World',
    state: 'FL',
    id: 2,
  },
];
const searchTerms = ['sarasota', 'fl'];
const lowerCasedTerms = searchTerms.map((term) => term.toLowerCase());
const params = ['city', 'state'];

const result = restaurants.filter((restaurant) =>
  lowerCasedTerms.every((lowerCasedTerm) =>
    params.some((param) => restaurant[param].toLowerCase().includes(lowerCasedTerm)),
  ),
);

console.log(result);