Question

给出这样的数组：

a = [{id:1, name:'a'}, {id:2, name:'b'}, {id:3, name:'c'}, {id:4, name:'d'}, {id:5, name:'e'}];

b = ['a','d'];

c = [1,4];

我如何从a和b获得c？我想从名字中获取id。谢谢你。

Answer 1

一种选择是使用.map和.find：

const a = [{id:1, name:'a'},{id:2, name:'b'}, {id:3, name:'c'},{id:4, name:'d'},{id:5, name:'e'}];
const b = ['a','d'];

const c = b.map(findName => 
  a.find(({ name }) => name === findName)
    .id
);
console.log(c);

您还可以将输入数组转换为按名称索引的对象，然后只对名称进行map转换，以便每次都不需要.find：

const a = [{id:1, name:'a'},{id:2, name:'b'}, {id:3, name:'c'},{id:4, name:'d'},{id:5, name:'e'}];
const b = ['a','d'];

const obj = a.reduce((accum, { id, name }) => {
  accum[name] = id;
  return accum;
}, {});
const c = b.map(name => obj[name]);
console.log(c);

Answer 2

b.forEach(itemB => {
  c.push(a.find(x=> x.name === itemB).id);
});

循环通过b数组并尝试find数组a中的c，如果存在，请将id推入man crontab.5

Answer 3

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main() {
int t;
cin>>t;
while((t--)>0){
    int n;
    cin>>n;
    int a[n];
    string s="";
    for(int i=0;i<n;i++){
        cin>>a[i];
        if(i==0){s+=to_string(a[i]); continue;}
        string s1 = s+to_string(a[i]); //sX
        string s2 = to_string(a[i])+s; //Xs

        if(stoi(s1)>=stoi(s2))s=s1;
        else s = s2;
    }
    cout<<s<<endl;
    }

return 0;
 }

Answer 4

你可以这样使用

var d = a.filter((item)=>{
   return b.indexOf(item.name) >= 0;
}); 
var c =  d.map((item) => {
  return item.id;
 });
// Result = [1,4]

Answer 5

请尝试这个

a = [ { id: 1, name: 'a' },
  { id: 2, name: 'b' },
  { id: 3, name: 'c' },
  { id: 4, name: 'd' },
  { id: 5, name: 'e' } ]
  
  b = ['a','d'];
  c = []
    a.map(res => { if (b.includes(res.name)) { c.push(res.id) } });
    alert(c)

如何在javascript中选择一些元素？

5 个答案: