我有这样的数组。
[{first_name: "john", age: "30", institution: {id: 1, name: "inst1"}},
{first_name: "john", age: "30", institution: {id: 2, name: "inst2"}},
{first_name: "john", age: "30", institution: {id: 3, name: "inst3"}}]
我想检查除机构外值是否相同，然后我想生成一个像这样的数组
[{first_name: "john", age: "30", institution:{name: "inst1, inst2, inst3"}}]
答案 0 :(得分：3)
如果您使用React.js
您可以尝试以下操作：
App.js
// App.js
import React from "react";
import { addItemToUser } from "./utiles";
export default class App extends React.Component {
state = {
users: [
{ first_name: "john", age: "30", institution: { id: 1, name: "inst1" } },
{ first_name: "mike", age: "20", institution: { id: 2, name: "inst2" } },
{ first_name: "ali", age: "40", institution: { id: 3, name: "inst3" } }
]
};
render() {
const newUser = {
first_name: "john",
age: "30",
institution: { id: 1, name: "inst2" }
};
const userList = addItemToUser(this.state.users, newUser);
console.log(userList);
return (
<div className="App">please check your console</div>
);
}
}
utiles.js
// utiles.js
export const addItemToUser = (users, userItemToAdd) => {
const existingUserItem = users.find(
user => user.first_name === userItemToAdd.first_name
);
if (existingUserItem) {
return users.map(user =>
user.first_name === userItemToAdd.first_name
? { ...user, institution: { name: user.institution.name + ', ' + userItemToAdd.institution.name } }
: user
);
}
return [...users, { ...userItemToAdd }];
};
答案 1 :(得分：1)
您要做的是某种分组，然后进行归约。 es6不提供开箱即用的groupBy功能。但是您可以使用reduce来实现相同的行为。
Source Array --> Group By Person --> Join the institution names
在您的情况下：
const arr = [{first_name: "john", age: "30", institution: {id: 1, name: "inst1"}},
{first_name: "john", age: "30", institution: {id: 2, name: "inst2"}},
{first_name: "john", age: "30", institution: {id: 3, name: "inst3"}}];
const result = arr.reduce((ac, person) => {
// check for the existing person in the accumulator
const existingRecord = ac.find(existingPerson => existingPerson.first_name === person.first_name)
// if. exists, concatenate the institution
if (existingRecord) {
existingRecord.institution += `, ${person.institution.name}`;
return ac;
// otherwise, add this as a new person
} else {
return [...ac, {
first_name: person.first_name,
age: person.age,
institution: person.institution.name
}]
}
}, []);
实施小组的更通用方法：
Most efficient method to groupby on an array of objects
您还可以使用 Lodash （提供更多选项）
答案 2 :(得分：0)
您可以尝试
const arr = [{first_name: "john", age: "30", institution: {id: 1, name: "inst1"}},
{first_name: "john", age: "30", institution: {id: 2, name: "inst2"}},
{first_name: "john", age: "30", institution: {id: 3, name: "inst3"}}];
const result = arr.reduce((acc, item) => {
const itemInAcc = acc.find(itemAcc => itemAcc.first_name === item.first_name && itemAcc.age === item.age);
if (!itemInAcc) {
acc.push({...item, institution: {name: item.institution.name}})
} else {
itemInAcc.institution.name = [itemInAcc.institution.name, item.institution.name].join(`, `)
}
return acc;
}, []);
console.log(result); // [{first_name: "john", age: "30", institution: {name: "inst1, inst2, inst3"}}]
答案 3 :(得分：0)
请使用以下代码。
let obj = [{
first_name: "john",
age: "30",
institution: {
id: 1,
name: "inst1"
}
},
{
first_name: "john",
age: "30",
institution: {
id: 2,
name: "inst2"
}
},
{
first_name: "john",
age: "30",
institution: {
id: 3,
name: "inst3"
}
}
]
let newObj = []
let isSame = true;
let obj1 = {}
let obj2 = {}
obj.forEach((o, i) => {
if (i + 1 < obj.length) {
obj1 = {
fn: obj[i].first_name,
age: obj[i].age
}
obj2 = {
fn: obj[i + 1].first_name,
age: obj[i + 1].age
}
if (JSON.stringify(obj1) !== JSON.stringify(obj2) ){
isSame = false;
}
}
})
if(isSame){
let anotherobj = {}
let fn = obj[0].first_name
let age = obj[0].age;
let instiValue=''
obj.forEach(o=>{
instiValue = instiValue + o.institution.name + ", ";
})
anotherobj.first_name = fn;
anotherobj.age = age;
anotherobj.institution = instiValue;
newObj.push(anotherobj);
}
console.log(newObj)
答案 4 :(得分：0)
这可能有帮助。
const data = [{first_name: "john", age: "30", institution: {id: 1, name: "inst1"}},
{first_name: "john", age: "30", institution: {id: 2, name: "inst2"}},
{first_name: "john", age: "30", institution: {id: 3, name: "inst3"}}]
console.log(
Object.values(
data.reduce((p,{ institution: { name }, ...r }, _, __, k = r) =>
((p[k] ? (p[k].institution.name += `, ${name}`)
: p[k] = ({ ...r, institution: { name } })), p),{})))
答案 5 :(得分：0)
您可以生成将first_name和age属性组合在一起的键，并在Map Object中使用它们，然后将结果转换回数组
const items = [
{first_name: "john", age: "30", institution: {id: 1, name: "inst1"}},
{first_name: "john", age: "30", institution: {id: 2, name: "inst2"}},
{first_name: "john", age: "30", institution: {id: 3, name: "inst3"}}];
const key = item => `${item.first_name}:${item.age}`;
const unique = new Map(items.map(item => [key(item), item]))
const result = Array.from(unique.values())
答案 6 :(得分：0)
请检查此代码-
let a = [{ first_name: "john", age: "30", institution: { id: 1, name: "inst1" } },
{ first_name: "john", age: "30", institution: { id: 2, name: "inst2" } },
{ first_name: "mohan", age: "20", institution: { id: 2, name: "inst2" } },
{ first_name: "john", age: "30", institution: { id: 3, name: "inst3" } }];
let b = [];
for (var i = 0; i < a.length; i++) {
if (b.length != 0) {
let j_val = '';
let flag = false;
for (let j = 0; j < b.length; j++) {
if ((a[i].first_name == b[j].first_name) && (a[i].age == b[j].age)) {
j_val = j;
flag = true;
} else {
}
}
let cc = a[i].institution.name;
if (flag) {
b[j_val].institution.name = b[j_val].institution.name + "," + cc;
} else {
b.push({ first_name: a[i].first_name, age: a[i].age, institution: { name: a[i].institution.name } });
}
} else {
b.push({ first_name: a[i].first_name, age: a[i].age, institution: { name: a[i].institution.name } });
}
}
console.log(b)