我刚刚开始使用公寓,我想开发一个共享秘密的电梯。然而,我的初步测试并不乐观。
我现在所处的地方是我有以下内容:
app/middleware/static_elevator.rb
require 'apartment/elevators/generic'
class StaticElevator < Apartment::Elevators::Generic
# @return {String} - The tenant to switch to
def parse_tenant_name(request)
# request is an instance of Rack::Request
'my_schema_name'
end
end
使用:
config/initializers/apartment.rb
require 'apartment/elevators/generic'
# require 'apartment/elevators/domain'
#require 'apartment/elevators/subdomain'
#
# Apartment Configuration
#
Apartment.configure do |config|
config.tenant_names = lambda { Account.pluck(:schema) }
end
Rails.application.config.middleware.use 'StaticElevator'
因此,实质上每次请求进入数据库时都使用my_schema_name作为模式。在我的控制器中,我有:
class V1::StaffController < ApplicationController
def index
@staff = Staff.only_employees
end
end
我的测试是:
require 'rails_helper'
RSpec.describe V1::StaffController, type: :controller do
context 'no query string' do
subject { get :index }
it 'gets only the employees' do
expect(Staff).to receive(:only_employees).and_return([])
subject
end
end
end
当我运行测试时,我得到:
Randomized with seed 63457
V1::StaffController
no query string
gets only the employees (FAILED - 1)
Failures:
1) V1::StaffController no query string gets only the employees
Failure/Error: expect(Staff).to receive(:only_employees).and_return([])
ActiveRecord::StatementInvalid:
PG::UndefinedTable: ERROR: relation "users" does not exist
LINE 5: WHERE a.attrelid = '"users"'::regclass
^
: SELECT a.attname, format_type(a.atttypid, a.atttypmod),
pg_get_expr(d.adbin, d.adrelid), a.attnotnull, a.atttypid, a.atttypmod
FROM pg_attribute a LEFT JOIN pg_attrdef d
ON a.attrelid = d.adrelid AND a.attnum = d.adnum
WHERE a.attrelid = '"users"'::regclass
AND a.attnum > 0 AND NOT a.attisdropped
ORDER BY a.attnum
# ./spec/controllers/v1/staff_controller_spec.rb:8:in `block (3 levels) in <top (required)>'
# ------------------
# --- Caused by: ---
# PG::UndefinedTable:
# ERROR: relation "users" does not exist
# LINE 5: WHERE a.attrelid = '"users"'::regclass
# ^
# ./spec/controllers/v1/staff_controller_spec.rb:8:in `block (3 levels) in <top (required)>'
Top 1 slowest examples (0.01605 seconds, 94.7% of total time):
V1::StaffController no query string gets only the employees
0.01605 seconds ./spec/controllers/v1/staff_controller_spec.rb:7
Finished in 0.01694 seconds (files took 4.37 seconds to load)
1 example, 1 failure
我告诉我架构没有运行。我对么?如果是这样,我做错了什么?
答案 0 :(得分:1)
看起来您的测试数据库尚未正确迁移。
运行后会发生什么
bundle exec rake db:migrate RAILS_ENV=test
另外:在您的示例中,您只返回parse_tenant_name中的静态字符串,这应该是动态的,例如,当您将承租人名称存储在request.session [:key]中时。
根据您的配置,此租户名称也应出现在Account.pluck(:schema)中。
答案 1 :(得分:0)
所以看起来即使你已经在数据库中有了架构,你也需要一个schema.rb来使用rspec。一旦我提供了一个db / schema.rb,突然一切都与世界和apartmant(以及它的伴随测试)一样顺利运行。
你可以至少部分地将这一点归结为操作员过度自信:)