javascript - 在对象javascript express中过滤嵌套数组
问题描述
考虑以下对象:
[
{
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
}
]
每个对象代表一个项目。一个项目有很多选项卡。
我只想返回至少一个选项卡中包含的项目 permissionsUserID 记录的用户的 ID。
因此,如果登录的用户的 ID 为 8,则这些是我想要获取的项目:
[
{
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
}
]
这就是我所做的过滤器:
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;
}
},
解决方案
这是一种方法
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;
})
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