尝试使用 javascript 从三个现有对象创建一个新对象

Question

我有三个对象，每个对象都是从 Django-rest 框架序列化的。每个对象都是配置文件中用户的数组。每个个人资料都需要一个 org_id 和电子邮件才有效。 “activeUsers”和“inactiveUsers”永远不会有重复，每个元素共享相同的 org_id，而 user_id 和电子邮件将始终不同。 “nonProfileUsers”是排除的所有其他配置文件
“activeUsers”和“inactiveUsers”org_id，注意 user_id 和电子邮件可以而且将会出现多次。

 activeUsers = [
    { "org_id": 1, "user_id": 9, "firstName": "Joj", "lastName": "Blonde", "email": "bop@test.ca", },
    {"org_id": 1, "user_id": 1, "firstName": "Tyler", "lastName": "Whitfield", "email": "tyler@atg.net", },
  ]

inactiveUsers = [
  {  "org_id": 1, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
  {"org_id": 1, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", }
]

//user that are in different profiles
nonProfileUsers = [
  { "org_id": 2, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
  {  "org_id": 3, "user_id": 6, "firstName": "weak", "lastName": "jdf", "email": "not@not.ca", },
  {  "org_id": 3, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
  {  "org_id": 4, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
  {  "org_id": 3, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
  {  "org_id": 2, "user_id": 5, "firstName": "test", "lastName": "test", "email": "tester@tester.ca", },
  { "org_id": 2, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
]

我正在使用 java-script，我需要做一些事情。缩小“nonProfileUsers”以删除重复的 user_id 和电子邮件组合，但至少保留一个 org_id（保留哪个无关紧要）

scaledNonProfileUsers = [
          { "org_id": 2, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
          {  "org_id": 3, "user_id": 6, "firstName": "weak", "lastName": "jdf", "email": "not@not.ca", },
          {  "org_id": 2, "user_id": 5, "firstName": "test", "lastName": "test", "email": "tester@tester.ca", },
          { "org_id": 2, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
        ]

合并“activeUsers”和“inactiveUsers” -我能够完成第 2 步

 combinedUsers= [
        { "org_id": 1, "user_id": 9, "firstName": "Joj", "lastName": "Blonde", "email": "bop@test.ca", },
        {"org_id": 1, "user_id": 1, "firstName": "Tyler", "lastName": "Whitfield", "email": "tyler@atg.net", },
        {  "org_id": 1, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
       {"org_id": 1, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", }
      ]

通过比较“scaledNonProfileUsers”与“combinedUsers”创建一个新对象。 “scaledNonProfileUsers”中而不是“combinedUsers”中的电子邮件将制作最终对象。

finalObj= [
            {  "org_id": 3, "user_id": 6, "firstName": "weak", "lastName": "jdf", "email": "not@not.ca", },
            {  "org_id": 2, "user_id": 5, "firstName": "test", "lastName": "test", "email": "tester@tester.ca", },
          ]

我真的很难完成第 1 步和第 3 步。

如果我只隔离电子邮件，我可以像这样制作一个数组

[
    0:"non@not.ca",
    1:"tester@tester.ca"
  ]

但我需要至少包含一个 org_id 和电子邮件。

到目前为止，这段代码一直是我的方法。

const temp_active = [...inactiveUsers, ...activeUsers,]// master list of both active/inactive

  const scaledNonProfileUsers= []//isolate email and make a list of emails
  for (var i = 0; i < nonProfileUsers?.length; i++) {
    scaledNonProfileUsers.push(nonProfileUsers[i]?.email)
  }

  const combinedUsers= []//isolate email and make a list of emails
  for (var i = 0; i < temp_active?.length; i++) {
   combinedUsers.push(temp_active[i]?.email)
  }

  let combinedUsers_filter = combinedUsers.filter((c, index) => {//filter duplicated emails
    return combinedUsers.indexOf(c) === index;
  });
  let scaledNonProfileUsers_filter = scaledNonProfileUsers.filter((c, index) => {//filter duplicate emails
    return scaledNonProfileUsers.indexOf(c) === index;
  });

  var finalObj = [];
  for (var i = 0; i < scaledNonProfileUsers_filter?.length; i++) {
    if (!eleContainsInArray(combinedUsers_filter, scaledNonProfileUsers_filter[i])) {
      finalObj.push(scaledNonProfileUsers_filter[i])
    }
  }
  function eleContainsInArray(arr, element) {
    if (arr != null && arr.length > 0) {
      for (var i = 0; i < arr.length; i++) {
        if (arr[i] == element)
          return true;
      }
    }
    return false;
  }

  console.log(finalObj)

任何人的帮助将不胜感激。

Answer 1

const activeUsers = [
    {
        org_id: 1,
        user_id: 9,
        firstName: 'Joj',
        lastName: 'Blonde',
        email: 'bop@test.ca',
    },
    {
        org_id: 1,
        user_id: 1,
        firstName: 'Tyler',
        lastName: 'Whitfield',
        email: 'tyler@atg.net',
    },
]

const inactiveUsers = [
    {
        org_id: 1,
        user_id: 3,
        firstName: 'James',
        lastName: 'Bond',
        email: 'test@test.ca',
    },
    {
        org_id: 1,
        user_id: 2,
        firstName: 'Chen',
        lastName: 'rain',
        email: 'posihdun@gmail.com',
    },
]

const nonProfileUsers = [
    {
        org_id: 2,
        user_id: 2,
        firstName: 'Chen',
        lastName: 'rain',
        email: 'posihdun@gmail.com',
    },
    {
        org_id: 3,
        user_id: 6,
        firstName: 'weak',
        lastName: 'jdf',
        email: 'not@not.ca',
    },
    {
        org_id: 3,
        user_id: 2,
        firstName: 'Chen',
        lastName: 'rain',
        email: 'posihdun@gmail.com',
    },
    {
        org_id: 4,
        user_id: 2,
        firstName: 'Chen',
        lastName: 'rain',
        email: 'posihdun@gmail.com',
    },
    {
        org_id: 3,
        user_id: 3,
        firstName: 'James',
        lastName: 'Bond',
        email: 'test@test.ca',
    },
    {
        org_id: 2,
        user_id: 5,
        firstName: 'test',
        lastName: 'test',
        email: 'tester@tester.ca',
    },
    {
        org_id: 2,
        user_id: 3,
        firstName: 'James',
        lastName: 'Bond',
        email: 'test@test.ca',
    },
]

const filteredNonProfileUsers = nonProfileUsers.reduce((prev, current) => {
    if (
        prev.find(
            user =>
                user.user_id === current.user_id || user.email === current.email
        )
    )
        return prev

    return [...prev, current]
}, [])

const activeUsersAndInactiveUsers = [...activeUsers, ...inactiveUsers]

const result = filteredNonProfileUsers.reduce((prev, current) => {
    if (
        activeUsersAndInactiveUsers.find(
            user =>
                user.user_id === current.user_id || user.email === current.email
        )
    )
        return prev

    return [...prev, current]
}, [])

console.log(result)

Answer 2

使用reduce()在数组内部循环，使用include()过滤元素匹配或不匹配：

var activeUsers = [
    { "org_id": 1, "user_id": 9, "firstName": "Joj", "lastName": "Blonde", "email": "bop@test.ca", },
    {"org_id": 1, "user_id": 1, "firstName": "Tyler", "lastName": "Whitfield", "email": "tyler@atg.net", },
  ]

var inactiveUsers = [
  {  "org_id": 1, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
  {"org_id": 1, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", }
]

//user that are in different profiles
var nonProfileUsers = [
  { "org_id": 2, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
  {  "org_id": 3, "user_id": 6, "firstName": "weak", "lastName": "jdf", "email": "not@not.ca", },
  {  "org_id": 3, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
  {  "org_id": 4, "user_id": 2, "firstName": "Chen", "lastName": "rain", "email": "posihdun@gmail.com", },
  {  "org_id": 3, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
  {  "org_id": 2, "user_id": 5, "firstName": "test", "lastName": "test", "email": "tester@tester.ca", },
  { "org_id": 2, "user_id": 3, "firstName": "James", "lastName": "Bond", "email": "test@test.ca", },
]

const scaledNonProfileUsers = nonProfileUsers.reduce((prev, curr) => (prev.map(el => el.user_id).includes(curr.user_id) || prev.map(el => el.email).includes(curr.email)) ? [...prev] : [...prev, curr], [])
const combinedUsers = [...inactiveUsers, ...activeUsers]
const finalObj =  scaledNonProfileUsers.reduce((prev, curr) => (combinedUsers.map(el => el.email).includes(curr.email)) ? [...prev] : [...prev, curr], [])
console.log(finalObj)

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