在 object javascript express 中过滤嵌套数组

Question

Considering the below object:

[
  {
    id: 5fc0be2990a8a12cc0ba0b5c,
    projectName: 'E-271120-B',
    projectManagaer: '5f7f1ba973ff621da4322248',
    dataInici: 2020-11-26T23:00:00.000Z,
    dataEntrega: 2020-11-26T23:00:00.000Z,
    dtoGlobal: null,
    dtoProjecte: null,
    archived: false,
    created: 2020-11-27T08:51:57.242Z,
    updated: 2021-01-25T10:01:18.733Z
    tabs: [{permissionsUserID:[250,8]},{permissionsUserID:[3]}],
    __v: 3
  },
  {
    tabs: [{permissionsUserID:[3,350]},{permissionsUserID:[15]}],
    _id: 5fc0be4690a8a12cc0ba0b5f,
    projectManagaer: '5f7f0e69b5862e1a085db388',
    projectName: 'E-271120-C',
    dataInici: 2020-11-27T23:00:00.000Z,
    dataEntrega: 2020-11-29T23:00:00.000Z,
    dtoGlobal: null,
    dtoProjecte: null,
    archived: false,
    created: 2020-01-21T08:46:41.958Z,
    updated: 2021-01-21T08:46:41.958Z,
    __v: 2
  },
  {
    tabs: [{permissionsUserID:[31,350]},{permissionsUserID:[8,893]}],
    _id: 5fc0be4690a8a12cc0ba0b5f,
    projectManagaer: '5f7f0e69b5862e1a085db388',
    projectName: 'E-23410-C',
    dataInici: 2020-11-27T23:00:00.000Z,
    dataEntrega: 2020-11-29T23:00:00.000Z,
    dtoGlobal: null,
    dtoProjecte: null,
    archived: false,
    created: 2020-01-21T08:46:41.958Z,
    updated: 2021-01-21T08:46:41.958Z,
    __v: 2
  }
]

Each object represents a Project. A project has many tabs.

I want to return only the projects that at least one tab contains in permissionsUserID the ID of the user that is logged.

So if the user that is logged has the ID 8, these are the projects I want to obtain:

[
  {
    id: 5fc0be2990a8a12cc0ba0b5c,
    projectName: 'E-271120-B',
    projectManagaer: '5f7f1ba973ff621da4322248',
    dataInici: 2020-11-26T23:00:00.000Z,
    dataEntrega: 2020-11-26T23:00:00.000Z,
    dtoGlobal: null,
    dtoProjecte: null,
    archived: false,
    created: 2020-11-27T08:51:57.242Z,
    updated: 2021-01-25T10:01:18.733Z
    tabs: [{permissionsUserID:[250,8]},{permissionsUserID:[3]}],
    __v: 3
  },
{
    tabs: [{permissionsUserID:[31,350]},{permissionsUserID:[8,893]}],
    _id: 5fc0be4690a8a12cc0ba0b5f,
    projectManagaer: '5f7f0e69b5862e1a085db388',
    projectName: 'E-23410-C',
    dataInici: 2020-11-27T23:00:00.000Z,
    dataEntrega: 2020-11-29T23:00:00.000Z,
    dtoGlobal: null,
    dtoProjecte: null,
    archived: false,
    created: 2020-01-21T08:46:41.958Z,
    updated: 2021-01-21T08:46:41.958Z,
    __v: 2
  }
]

That's the filter I have done:

async getAll(pagination, user) {
    try {
      const filter = {};
      if(pagination.archived) {
        filter['archived'] = pagination.archived;
      }
      if(pagination.search) {
        filter['$text'] = {$search: pagination.search}
      }

      const { Project: projectSchema } = this.getSchemas();
    

      const projectsDocs = await projectSchema.paginate(filter, {
        limit: pagination.limit ? parseInt(pagination.limit) : 10,
        page: pagination.page ? parseInt(pagination.page) + 1 : 1
      });

      if (!projectsDocs) {
        throw new errors.NotFound('No Projects.');
      }

      projectsDocs.docs.forEach(element => {
        element.tabs.filter( d => d.permissionsUserID.every( c => c.includes(user._id)));
      });

      return projectsDocs;
    } catch (error) {
      throw error;
    }
},

Answer 1

Here is one way

const data = [...];
const userId = 8;
const result = data.filter((item) => {
    const {tabs} = item;
    let loggedIn = false;
    tabs.forEach((tab) => {
        if (tab.permissionsUserID.includes(userId)) {
            loggedIn = true;
            return true
        }
    })
    return loggedIn;
})

Answer 2

Here's a simple function which should get you what you want.

Filter() returns a subset of the projects list. Some() returns true if at least one of the tabs has the value we're looking for. Includes() returns true if the permissionsUserId list has the user id we want. Chain those together and you get the subset of projects where a tab's permissions has the desired user id.

const data = [
        /* list of projects */
    ],
    userId = 8;

function getProjectsForUserId (data, userId) {
    return data.filter((project) => {
        return project.tabs.some((tab) => {
            return tab.permissionsUserID.includes(userId);
        });
    });
}

console.log(getProjectsForUserId(data, 8));