Mongoose 查找 array.length 大于 0 & 的所有文档对数据进行排序 [英] Mongoose find all documents where array.length is greater than 0 & sort the data

问题描述

我正在使用 mongoose 在 MongoDB 上执行 CRUD 操作.这就是我的架构的外观.

I am using mongoose to perform CRUD operation on MongoDB. This is how my schema looks.

var EmployeeSchema = new Schema({
      name: String,
      description: {
        type: String,
        default: 'No description'
      },
      departments: []

    });

每个员工可以属于多个部门.Departments 数组看起来像 [1,2,3].在这种情况下，departments.length = 3.如果员工不属于任何部门，则departments.length 将等于0.

Each employee can belong to multiple department. Departments array will look like [1,2,3]. In this case departments.length = 3. If the employee does not belong to any department, the departments.length will be equal to 0.

我需要找到 EmployeeSchema.departments.length > 0 & 的所有员工如果查询返回超过 10 条记录，我只需要获取最大部门数的员工.

I need to find all employee where EmployeeSchema.departments.length > 0 & if query return more than 10 records, I need to get only employees having maximum no of departments.

是否可以使用 Mongoose.find() 来获得想要的结果?

Is it possible to use Mongoose.find() to get the desired result?

推荐答案

假设您的模型名为 Employee:

Employee.find({ "departments.0": { "$exists": true } },function(err,docs) {

})

作为 $exists 请求数组的 0 索引，这意味着它里面有东西.

As $exists asks for the 0 index of an array which means it has something in it.

同样适用于最大数量:

Employee.find({ "departments.9": { "$exists": true } },function(err,docs) {

})

因此需要在数组中至少有 10 个条目才能匹配.

So that needs to have at least 10 entries in the array to match.

实际上，您应该记录数组的长度并在每次添加内容时使用 $inc 进行更新.然后你可以这样做:

Really though you should record the length of the array and update with $inc every time something is added. Then you can do:

Employee.find({ "departmentsLength": { "$gt": 0 } },function(err,docs) {

})

在您存储的departmentsLength"属性上.该属性可以编入索引，从而提高效率.

On the "departmentsLength" property you store. That property can be indexed, which makes it much more efficient.

这篇关于Mongoose 查找 array.length 大于 0 & 的所有文档对数据进行排序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！