更改过滤数组中的布尔值

Question

我有一个数组，我需要编写一个接受用户名的函数，并且布尔值应该更改为 true。

var array = [{
  user: "Sasha",
  message: "Hello guys",
  time: "20:28:2",
  read: false
}, {
  user: "Sasha",
  message: "How are you doing",
  time: "20:28:2",
  read: false
}, {
  user: "Dima",
  message: "I am fine, thanks!",
  time: "20:28:2",
  read: false
}, {
  user: "Katya",
  message: "I am doing well! What about you?",
  time: "20:28:2",
  read: false
}]

function readMessage(user) {
  let test = array
  let filtered = test.filter(item => item.user === user);
  let y = filtered.map(item => item.user && !item.read);
  console.log(y);
}

readMessage();

我认为我应该过滤数组，然后将 bool 改为相反，但之后 map 函数只返回 bool。如何更改布尔值并将更改推送到原始数组？

Answer 1

首先你需要用你想要的值过滤数组，然后你得到一个过滤后的数组，现在你映射这个数组来更新布尔值。

// put your messages array in a variable
var array = [{
  user: "Sasha",
  message: "Hello guys",
  time: "20:28:2",
  read: false
}, {
  user: "Sasha",
  message: "How are you doing",
  time: "20:28:2",
  read: false
}, {
  user: "Dima",
  message: "I am fine, thanks!",
  time: "20:28:2",
  read: false
}, {
  user: "Katya",
  message: "I am doing well! What about you?",
  time: "20:28:2",
  read: false
}]

function readMessage(user) {
  let test = array
  let filtered = test.filter(item => item.user === user);
  // console.log(filtered); // check the filtered array;
  filtered.map(item => item.read = true);
  console.log(filtered);
}
// call the function now;
readMessage('Sasha');

Answer 2

这行代码有效但只是返回您过滤/映射的值：

array.filter(item => item.user === 'Sasha').map(item => item.read = true)

在你的函数中返回数组

array.filter(item => item.user === user).map(item => item.read = true)
return array;

Answer 3

您可以像这样在 map 函数中将 read 设置为 true：

function readMessage(user){
   let test = array
   let filtered = test.filter(item => item.user === user);
   let y = filtered.map(item => item.read = true);
}

[]的

Answer 4

此版本不会就地修改您的消息，而只会返回过滤后的更新列表：

const items = [
  {user: "Sasha", message: "Hello guys", time: "20:28:2", read: false},
  {user: "Sasha", message: "How are you doing", time: "20:28:2", read: false},
  {user: "Dima", message: "I am fine, thanks!", time: "20:28:2", read: false},
  {user: "Katya", message: "I am doing well! What about you?", time: "20:28:2", read: false}
] 

const readMessages = (items, user) => items
  .filter(item => item.user === user)
  .map(({read, ...rest}) => ({...rest, read: !read}))

const updatedItems = readMessages(items, 'Sasha')

console.log(updatedItems)

此版本返回一个新的完整消息列表，匹配的消息已更改：

const items = [
  {user: "Sasha", message: "Hello guys", time: "20:28:2", read: false},
  {user: "Sasha", message: "How are you doing", time: "20:28:2", read: false},
  {user: "Dima", message: "I am fine, thanks!", time: "20:28:2", read: false},
  {user: "Katya", message: "I am doing well! What about you?", time: "20:28:2", read: false}
] 

const updatedMessages = (items, userToCheck) => items
  .map(({user, read, ...rest}) => user == userToCheck
      ? {...rest, user, read: !read}
      : {...rest, user, read}
  )

const updatedItems = updatedMessages(items, 'Sasha')

console.log(updatedItems)

而且这个版本更新了项目列表。 （除非绝对必要，否则我建议不要这样做。不可变数据有很大的好处！）

const updateMessages = (items, userToCheck) => {
  items.forEach(item => {
    if (item.user == userToCheck) {
      item.read = !item.read
    }
  })
  return items
}