Question

JavaScript：

exports.userWithoutProject = async (req, res, next) => {
  try {
    let benchUser = [];
    const project = await Project.find({})
    const users = await User.find({}, {
      email: 1,
      _id: 0
    })

    for (let i = 0; i < users.length; i++) {
      for (let j = 0; j < project.length; j++) {
        for (let k = 0; k < project[j].seats.length; k++) {
          if (users[i].email !== project[j].seats[k].employee && project[j].seats[k].employee === undefined) {
            benchUser.push(users[i].email)
          }
        }
      }
    }

    const uniq = [...new Set(benchUser)];

    return res.json({
      users: uniq
    })
  } catch (error) {
    next(error)
  }
}

用户：

[{ 
   email: 'user1@email.com',
 }, { 
   email: 'user2@email.com',
 }, {
   email: 'user3@email.com' 
 }]

项目（对象数组）：

{ _id: 5cc2dd2eb3eea7004c9a7240,
  name: 'Project One',
  description:
   'Lorem Ipsum',
  start: '2018-06-01T09:45:00.000Z',
  end: '2019-12-31T09:45:00.000Z',
  seats:
   [ 
     { 
       skills: [Array],
       _id: 5cc2e3cab3eea7004c9a724a,
       start: '2018-06-01T09:45:00.000Z',
       end: '2019-12-31T09:45:00.000Z',
       potentialExtension: '2020-06-31T09:45:00.000Z',
       role: 'Dev',
       approved: true,
       workload: 10,
       employee: 'user1@email.com' 
     },
     { 
       skills: [Array],
       _id: 5cc2e3cab3eea7004c9a7241,
       start: '2018-06-01T09:45:00.000Z',
       end: '2019-12-31T09:45:00.000Z',
       potentialExtension: '2020-06-31T09:45:00.000Z',
       role: 'PM',
       approved: true,
       workload: 20,
       employee: 'user3@email.com' 
     }
   ]
}

我的目标：

我想让所有没有项目席位的用户退回。

我有多个项目，每个项目都有一个seat，其中用户电子邮件写在employee字段中。因此，我必须浏览所有项目，并遍历每个项目席位，以检查我的用户是否匹配。

我从上面的方法返回user2@email.com正确，但返回user1@email.com却不正确。

Answer 1

您可以使用Array.prototype.filter和Array.prototype.includes

来满足您的要求。

let users = [{ 
	email: 'user1@email.com',
}, { 
	email: 'user2@email.com',
}, {
	email: 'user3@email.com' 
}];

let projects = [{ _id: '5cc2dd2eb3eea7004c9a7240',
  name: 'Project One',
  description:
   'Lorem Ipsum',
  start: '2018-06-01T09:45:00.000Z',
  end: '2019-12-31T09:45:00.000Z',
  seats:
   [ 
     { 
       skills: [Array],
       _id: '5cc2e3cab3eea7004c9a724a',
       start: '2018-06-01T09:45:00.000Z',
       end: '2019-12-31T09:45:00.000Z',
       potentialExtension: '2020-06-31T09:45:00.000Z',
       role: 'Dev',
       approved: true,
       workload: 10,
       employee: 'user1@email.com' 
     },
     { 
       skills: [Array],
       _id: '5cc2e3cab3eea7004c9a7241',
       start: '2018-06-01T09:45:00.000Z',
       end: '2019-12-31T09:45:00.000Z',
       potentialExtension: '2020-06-31T09:45:00.000Z',
       role: 'PM',
       approved: true,
       workload: 20,
       employee: 'user3@email.com' 
     }
   ]
}];

let projectUsers = new Set();
projects.forEach(({seats}) => {
	seats.forEach(({employee}) => projectUsers.add(employee));
});

let out = users.filter(({email}) => !Array.from(projectUsers).includes(email));
console.log(out)

MongoDB仅返回来自其他文档的用户与字段不匹配的文档

1 个答案: