Question

I have model:

# encoding: utf-8

class Tag
    include Mongoid::Document
    field :name, type: String
    field :count, type: Integer
    index :name, unique: true

    validates_uniqueness_of :name

    def self.create_tag(name)
        tag = Tag.new
        tag.name = name
        tag.count = 0
        tag.save
    end

    def self.find_by_name(name)
        Tag.where(name: name).entries
    end
end

And I have test for the model:

describe Tag, "# simple database operations" do
    it " - insert test records" do
        Tag.create_tag("joe")
        Tag.create_tag("joe")

        p Tag.find_by_name("joe")
    end
end

If I look at collection after test execution I’ll find only one record, but I want to catch exception in the case of duplicate record insertion.

Is it possible?

Answer 1

By default Mongoid writes in “fire and forget” mode. It sends a write and returns immediately. To check for error, you should write in “safe mode”. Try this.

def self.create_tag(name)
    tag = Tag.new
    tag.name = name
    tag.count = 0
    tag.safely.save! # <= note the 'safely' here. Also bang version of save is used.
end

Or, better yet,

def self.create_tag(name)
    Tag.safely.create!(name: name, count: 0)
end

See the doc here.