问题
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: []
});
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.
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.
Is it possible to use Mongoose.find() to get the desired result?
回答1:
Presuming your model is called Employee:
Employee.find({ "departments.0": { "$exists": true } },function(err,docs) {
})
As $exists asks for the 0 index of an array which means it has something in it.
The same applies to a maximum number:
Employee.find({ "departments.9": { "$exists": true } },function(err,docs) {
})
So that needs to have at least 10 entries in the array to match.
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) {
})
On the "departmentsLength" property you store. That property can be indexed, which makes it much more efficient.
回答2:
By some reason, selected answer doesn't work as for now. There is the $size operator.
Usage:
collection.find({ field: { $size: 1 } });
Will look for arrays with length 1.
回答3:
use can using $where like this:
await EmployeeSchema.find( {$where:'this.departments.length>0'} )
来源：https://stackoverflow.com/questions/31677061/mongoose-find-all-documents-where-array-length-is-greater-than-0-sort-the-data