打字稿过滤对象数组

Question

我想results根据存储在过滤器中的值过滤我的数组。我试过下面的代码。但它不适用。

let filters = {
  name: ["Krishna", "Naveen"],
  city : ["London"]
};
results = [
{
    "name": "Krishna#Surname",
    "city": "London",
    "age": 23
},
{
    "name": "Naveen#Surname",
    "city": "London",
    "age": 23
},
{
    "name": "Krishna#Surname",
    "city": "NewYork",
    "age": 23
},
{
    "name": "Praveen#Surname",
    "city": "Washington",
    "age": 23
}
]
this.results1 = this.multiFilter(results,filters);

multiFilter(array:any=[], filters:Object) {
const filterKeys = Object.keys(filters);
return array.filter((item) => {
  return filterKeys.every(key => {
    let filters1= filters[key];
    return filters1.every(key1 => {
      return !!~ item[key].indexOf(key1)
    });
  });
});
}

Answer 1

另一种方法是将函数filter与函数includes和一起使用some。

let filters = { name: ["Krishna", "Naveen"], city : ["London"]},
    results = [{    "name": "Krishna#Surname",    "city": "London",    "age": 23},{    "name": "Naveen",    "city": "London",    "age": 23},{    "name": "Krishna",    "city": "NewYork",    "age": 23},{    "name": "Praveen",    "city": "Washington",    "age": 23}],
    result = results.filter(({name, city}) => filters.name.some(n => name.includes(n)) && filters.city.includes(city));

console.log(result);

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Answer 2

您可以检查每个过滤器并返回匹配所有条件的对象。

这种方法适用于任意数量的条件。

function multiFilter(array, filters) {
    return array.filter(o =>
        Object.keys(filters).every(k =>
            [].concat(filters[k]).some(v => o[k].includes(v))));
}

var filters = { name: ["Krishna", "Naveen"], city: ["London"] },
    results = [{ name: "Krishna#Surname", city: "London", age: 23 }, { name: "Naveen#Surname", city: "London", age: 23 }, { name: "Krishna#Surname", city: "NewYork", age: 23 }, { name: "Praveen#Surname", city: "Washington", age: 23 }],
    filtered = multiFilter(results, filters);
  
  console.log(filtered);

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