我有一个对象数组
people = [
{id: "1", name: "abc", gender: "m", age:"15" },
{id: "2", name: "a", gender: "m", age:"25" },
{id: "3", name: "efg", gender: "f", age:"5" },
{id: "4", name: "hjk", gender: "m", age:"35" },
{id: "5", name: "ikly", gender: "m", age:"41" },
{id: "6", name: "ert", gender: "f", age:" 30" },
{id: "7", name: "qwe", gender: "f", age:" 31" },
{id: "8", name: "bdd", gender: "m", age:" 78" },
]
我有另一个阵列：
id_filter = [1,4,5,8]
如果id与id_filter数组和gender = m
如何使用过滤功能在javascript中执行此操作？
答案 0 :(得分：8)
您可以使用 array.filter() ，条件很少，无法获得所需的输出。我也纠正了你的JSON。
var filtered = people.filter(function(item) {
return id_filter.indexOf(item.id) !== -1 && item.gender==='m';
});
<强>样本
var people =[
{ "id": 1, "name": "abc", "gender": "m","age": "15" },
{ "id": 2, "name": "a", "gender": "m", "age": "25" },
{ "id": 3,"name": "efg", "gender": "f","age": "5" },
{ "id": 4,"name": "hjk","gender": "m","age": "35" },
{ "id": 5, "name": "ikly","gender": "m","age": "41" },
{ "id": 6, "name": "ert", "gender": "f", "age": "30" },
{ "id": 7, "name": "qwe", "gender": "f", "age": "31" },
{ "id":8, "name": "bdd", "gender": "m", "age": " 8" }
];
var id_filter = [1,4,5,8];
var filtered = people.filter(function(item) {
return id_filter.indexOf(item.id) !== -1 && item.gender==='m';
});
console.log(filtered);
答案 1 :(得分：4)
使用Array.includes()功能：
var people = [
{id : "1", name : "abc", gender : "m", age :"15" }, {id : "2", name : "a", gender : "m", age :"25" },
{id : "3", name : "efg", gender : "f", age :"5" }, {id : "4", name : "hjk", gender : "m", age :"35" },
{id : "5", name : "ikly", gender : "m", age :"41" }, {id : "6", name : "ert", gender : "f", age :" 30" },
{id : "7", name : "qwe", gender : "f", age :" 31" }, {id : "8", name : "bdd", gender : "m", age :" 78" }
],
id_filter = [1,4,5,8],
result = people.filter((o) => id_filter.includes(+o.id) && o.gender == "m");
console.log(result);
+o.id - 此处使用+将数字字符串转换为数字答案 2 :(得分：1)
最简单的过滤方法是使用javascript filter函数，类似于：
people.filter((person) => id_filter.includes(person.id))
答案 3 :(得分：1)
在这种情况下，您可以使用filter和include函数，因为您的id是字符串，因此在使用前需要进行解析。
var result = people.filter((person) => (id_filter.includes(parseInt(person.id)) && person.gender ==='m'))
答案 4 :(得分：0)
在这种情况下，这样做更有意义
所以首先你需要遍历你的people数组
然后你检查一个人ID是否等于你的过滤列表
for(person in people) {
for(id in id_filter) {
if(person[id] == id && person[gender] == "m"){
}
}
}
答案 5 :(得分：0)
您可以使用Array.prototype.filter之类的：
function filter(arr, ids, gender) { // takes an array of people arr, an array of indexes ids, and a gender and return the matched people objects from arr
return arr.filter(function(obj) { // filtering each object...
return ids.includes(obj.id) && obj.gender === gender; // if this object is is included in the ids array and if its gender property is equal to the desired gender
});
}
var people = [{id:"1",name:"abc",gender:"m",age:"15"},{id:"2",name:"a",gender:"m",age:"25"},{id:"3",name:"efg",gender:"f",age:"5"},{id:"4",name:"hjk",gender:"m",age:"35"},{id:"5",name:"ikly",gender:"m",age:"41"},{id:"6",name:"ert",gender:"f",age:"30"},{id:"7",name:"qwe",gender:"f",age:"31"},{id:"8",name:"bdd",gender:"m",age:"78"}];
console.log(filter(people, ["5", "7", "4"], "m")); // filtering elements where id is one of ["5", "7", "4"] and the gender is "m".
注意： id中对象的people属性是字符串，因此您必须向filter提供一个字符串ID数组或转换为id属性到一个数字，然后再传递给includes。
答案 6 :(得分：0)
const people = [
{id: "1", name: "abc", gender: "m", age: "15" },
{id: "2", name: "a", gender: "m", age: "25" },
{id: "3", name: "efg", gender: "f", age: "5" },
{id: "4", name: "hjk", gender: "m", age: "35" },
{id: "5", name: "ikly", gender: "m", age: "41" },
{id: "6", name: "ert", gender: "f", age: " 30" },
{id: "7", name: "qwe", gender: "f", age: " 31" },
{id: "8", name: "bdd", gender: "m", age: " 78" },
]
const idFilter = [1,4,5,8]
const idIsInList = id => idFilter.includes(+id) // "+id" to make sure it is a number, not a string
const genderIsMale = gender => gender === "m"
const result = people.filter(item => idIsInList(item.id) && genderIsMale(item.gender))
console.log(result)
答案 7 :(得分：0)
您可以通过以下代码实现：
var filtered_people = people.filter(function(person){
return id_filter.includes(person.id) && person.gender === 'm';
});
确保每个人的id都是整数，而不是字符串，如示例所示。否则，includes（）函数不匹配。此外，您的people数组存在内部语法问题。所以，最终的代码看起来像这样：
var people = [
{id: 1, name: "abc", gender: "m", age:15},
{id: 2, name: "a", gender: "m", age: 25},
{id: 3, name: "efg", gender: "f", age: 5},
{id: 4, name: "hjk", gender: "f", age: 35},
{id: 5, name: "ikly", gender: "m", age: 41},
{id: 6, name: "ert", gender: "f", age: 30},
{id: 7, name: "qwe", gender: "f", age: 31},
{id: 8, name: "bdd", gender: "m", age: 78}
];
var id_filter = [1,4,5,8];
var filtered_people = people.filter(function(person){
return id_filter.includes(person.id) && person.gender === 'm';
});
我希望这会对你有所帮助。 祝你好运。
答案 8 :(得分：0)
如果您的id_filter大，则需要先将其转换为new Set()。这将允许进行恒定时间查找。然后，您可以使用people迭代.filter()数组，如果其中的集合.has()中的true且性别等于{{1 }}。
id
总体而言，此方法的时间复杂度为'm'，与使用const people = [ {id: "1", name: "abc", gender: "m", age:"15" }, {id: "2", name: "a", gender: "m", age:"25" }, {id: "3", name: "efg", gender: "f", age:"5" }, {id: "4", name: "hjk", gender: "m", age:"35" }, {id: "5", name: "ikly", gender: "m", age:"41" }, {id: "6", name: "ert", gender: "f", age:" 30" }, {id: "7", name: "qwe", gender: "f", age:" 31" }, {id: "8", name: "bdd", gender: "m", age:" 78" }, ];
const id_filter = new Set([1,4,5,8]);
const res = people.filter(({id, gender}) => id_filter.has(+id) && gender === 'm');
console.log(res);或O(N)方法时得到的O(Nk)相反，其中{ {1}}是.includes()数组的长度，而.indexOf()是N数组的长度
答案 9 :(得分：-1)
people = [
{id : "1", name : "abc", gender : "m", age :"15" },
{id : "2", name : "a", gender : "m", age :"25" },
{id : "3", name : "efg", gender : "f", age :"5" },
{id : "4", name : "hjk", gender : "m", age :"35" },
{id : "5", name : "ikly", gender : "m", age :"41" },
{id : "6", name : "ert", gender : "f", age :" 30" },
{id : "7", name : "qwe", gender : "f", age :" 31" },
{id : "8", name : "bdd", gender : "m", age :" 78" }
]
var id_filter = ["1","4","5","8"], filteredPeople = [];
for( var i=people.length-1; i>=0; --i){
if( id_filter.indexOf( people[i].id ) != -1 ){
filteredPeople.push( people[i] );
}
}
console.log( filteredPeople );