使用Typescript过滤对象数组

Question

我想根据筛选器中存储的值来过滤我的 results 数组。我尝试了下面的代码。但它并没有起作用。

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)
    });
  });
});
}

- Krishna

Answer 1

你可以检查每个过滤器，返回符合所有条件的对象。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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- Nina Scholz

---

我已经做了一个小改变。我的名字包含#姓, 但在过滤器中只有克里希纳。我该如何进行筛选？ - Krishna

我无法弄清楚如何在我的情况下使用您的解决方案。您能否在这里看一下？https://stackoverflow.com/questions/62290308/using-filter-function-in-typescript - user13101751

Answer 2

另一种方法是使用filter函数与includes和some函数一起使用。

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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- Ele

---

我进行了一个小更改。我的名字包含#姓氏，但在过滤器中我只有Krishna。该如何进行筛选？ - Krishna

@Krishna，你是否也想接受那种类型的名称？ - Ele

是的，@Ele。我也需要过滤那些类型的值。 - Krishna

Answer 3

您可以使用filter函数过滤数组，并使用includes函数检查名称和城市是否在该数组中。

let filters = {name: ["Krishna", "Naveen"],city: ["London"]};
let results = [{"name": "Krishna","city": "London","age": 23},{"name": "Naveen","city": "London","age": 23},{"name": "Krishna","city": "NewYork","age": 23},{"name": "Praveen","city": "Washington","age": 23}]

let result = results.filter(o => filters.name.includes(o.name) && filters.city.includes(o.city));

console.log(result);

- Eddie