Question

FoundUser是id为用户将新项目添加到该特定用户项目数组的用户。

const FoundUser = Object.keys(users)
  .map(key => users[key])
  .find(item => item.uid === uid)
console.log(FoundUser, 'current obj user')

UpUser是更新后的FoundUser对象，其中的项目已添加到特定用户项目数组中。

const itemobj = { name: "addeditem", price:330 }
const UpUser = {...FoundUser, items:FoundUser.items.concat([itemobj])}
console.log(UpUser, 'UpUser')

如何在不删除唯一键users的情况下，仅在原始"97gbLdr4yyZljL75aZeVyoJraPo2"对象中更新同一用户的对象。

const final = Object.assign({}, )
console.log(final, 'final')

用户数据如下所示。

state= {
{ "97gbLdr4yyZljL75aZeVyoJraPo2": 
  {
   email:"some email", 
   items:[{name:'rose sherpa', price: 250}], 
   uid:'97gbLdr4yyZljL75aZeVyoJraPo2',
   username:'someusername'
  },
  "47gbLdr412312312yyZljL75o2": 
  {
   email:"some email2", 
   items:[{name:'black sherpa', price: 350}], 
   uid:'97gbLdr4yy1231123VyoJraPo2',
   username:'someusername2'
  } 
 }
}

所需的输出

{"97gbLdr4yyZljL75aZeVyoJraPo2": 
 {
  email:"some email", 
  items:[{name:'rose sherpa', price: 250} , {name:'addeditem', price:330}], 
  uid:'97gbLdr4yyZljL75aZeVyoJraPo2',
  username:'someusername'
 },
 "47gbLdr412312312yyZljL75o2": 
 {
  email:"some email2", 
  items:[{name:'black sherpa', price: 350}], 
  uid:'97gbLdr4yy1231123VyoJraPo2',
  username:'someusername2'
 } 
}

Answer 1

与您更新items FoundUser的方式类似：

const final = { ...users, [FoundUser.uid]: UpUser }

顺便说一句Object.keys(users).map(key => users[key])可以简化为Object.values(users)。

Answer 2

我可能将所有内容都折叠成一个函数：

const users = {"47gbLdr412312312yyZljL75o2": {"email": "some email2", "items": [{"name": "black sherpa", "price": 350}], "uid": "97gbLdr4yy1231123VyoJraPo2", "username": "someusername2"}, "97gbLdr4yyZljL75aZeVyoJraPo2": {"email": "some email", "items": [{"name": "rose sherpa", "price": 250}], "uid": "97gbLdr4yyZljL75aZeVyoJraPo2", "username": "someusername"}}

const addItem = (item, uid, users) => 
  Object.values(users).find(user => user.uid === uid).items.push(item)


addItem({name:'addeditem', price:330}, '97gbLdr4yyZljL75aZeVyoJraPo2', users)

console.log(users)

或者如果您不能保证您想要的用户将被包含在列表中，那么

const addItem = (item, uid, users) => {
  const user = Object.values(users).find(user => user.uid === uid)
  if (user) {user.items.push(item)}
  return users
}

（当然，如果user.items可能不存在，可能会有另一项检查。）

我实际上偏好不可变数据，并且我自己可能会使用Ramda制作解决方案。（免责声明：我是Ramda的作者）

const {map, when, propEq, evolve, append} = R

const users = {"47gbLdr412312312yyZljL75o2": {"email": "some email2", "items": [{"name": "black sherpa", "price": 350}], "uid": "97gbLdr4yy1231123VyoJraPo2", "username": "someusername2"}, "97gbLdr4yyZljL75aZeVyoJraPo2": {"email": "some email", "items": [{"name": "rose sherpa", "price": 250}], "uid": "97gbLdr4yyZljL75aZeVyoJraPo2", "username": "someusername"}}


const addItem = (item, uid, users) => 
  map(when(propEq('uid', uid), evolve({items: append(item)})), users)


const newUsers = addItem({name:'addeditem', price:330}, '97gbLdr4yyZljL75aZeVyoJraPo2', users)

console.log(newUsers)
console.log(users) // unchanged

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>

在对象中用id替换对象，并在数据库中保持相同的键

2 个答案: