Mongoose unique if not null and if state - node.js

Question

I have a unique index like this
code: {
type: String,
index: {
unique: true,
partialFilterExpression: {
code: { $type: 'string' }
}
},
default: null
},
state: { type: Number, default: 0 },
but When the state is 2 (archived) I want to keep the code, but it should be able to reuse the code, so it cannot be unique if state is 2.
Is there any away that I could accomplish this?

Answer 1

This is possible, though it's through a work around documented here https://jira.mongodb.org/browse/SERVER-25023.
In MongoDB 4.7 you will be able to apply different index options to the same field but for now you can add a non-existent field to separate the two indexes.
Here's an example using the work around.
(async () => {
const ItemSchema = mongoose.Schema({
code: {
type: String,
default: null
},
state: {
type: Number,
default: 0,
},
});
// Define a unique index for active items
ItemSchema.index({code: 1}, {
name: 'code_1_unique',
partialFilterExpression: {
$and: [
{code: {$type: 'string'}},
{state: {$eq: 0}}
]
},
unique: true
})
// Defined a non-unique index for non-active items
ItemSchema.index({code: 1, nonExistantField: 1}, {
name: 'code_1_nonunique',
partialFilterExpression: {
$and: [
{code: {$type: 'string'}},
{state: {$eq: 2}}
]
},
})
const Item = mongoose.model('Item', ItemSchema)
await mongoose.connect('mongodb://localhost:27017/so-unique-compound-indexes')
// Drop the collection for test to run correctly
await Item.deleteMany({})
// Successfully create an item
console.log('\nCreating a unique item')
const itemA = await Item.create({code: 'abc'});
// Throws error when trying to create with the same code
await Item.create({code: 'abc'})
.catch(err => {console.log('\nThrowing a duplicate error when creating with the same code')})
// Change the active code
console.log('\nChanging item state to 2')
itemA.state = 2;
await itemA.save();
// Successfully created a new doc with sama code
await Item.create({code: 'abc'})
.then(() => console.log('\nSuccessfully created a new doc with sama code'))
.catch(() => console.log('\nThrowing a duplicate error'));
// Throws error when trying to create with the same code
Item.create({code: 'abc'})
.catch(err => {console.log('\nThrowing a duplicate error when creating with the same code again')})
})();

Answer 2

This is not possible with using indexes. Even if you use a compound index for code and state there will still be a case where
new document
{
code: 'abc',
state: 0
}
archived document
{
code: 'abc',
state: 2
}
Now although you have the same code you will not be able to archive the new document or unarchive the archived document.
You can do something like this
const checkCode = await this.Model.findOne({code:'abc', active:0})
if(checkCode){
throw new Error('Code has to be unique')
}
else{
.....do something
}