What's the best way to retrieve today's tasks from Asana? - asana

I am trying to pull up a page that shows all the tasks that are assigned to me and that are marked as today.
gem 'asana', git: 'https://github.com/phcyso/asana.git'
//assume w is a valid workspace
#tasks1 = w.tasks(Asana::User.me.id)
<% #tasks.each do |t2| %>
<% if t2.assignee_status == "today" and t2.completed == false %>
<li class="list-group-item"><%=t2.name%></li>
<% end %>
<% end %>
The issue is that this is waaaaaay to slow since it iterates through my entire task list to find what's undone and marked as today. Is there a better way to do this? Couldn't find any API that would give me that data pre-filtered.
Thanks

You cannot currently filter on the server based on all of your criteria, but you can filter out completed tasks by specifying the completed_since=now query parameter. This will indicate it should only return tasks that have been completed after the current time, i.e. no complete tasks.
See https://asana.com/developers/api-reference/tasks#query for more options.
Assuming completed tasks are the majority of your assigned tasks, that should help considerably.

Related

Rails: Global instance variables in Rails

In my app users can submit recipes through a form, which will be published on a website. Before recipes get published they are moderated through a moderator.
Therefore my app shows in the navbar a count of all currently unpublished recipes for the moderator like so:
To achieve this at the moment I do the following:
application.rb
before_action :count_unpublished
def count_unpublished
#unpublished_count = Recipe.where(:published => false).count
end
_navbar.html.erb
<li>
<%= link_to recipes_path do %>
Recipes <span class="badge" style="background-color:#ff7373"><%= #unpublished_count %></span>
<% end %>
</li>
It works, but I am wondering now if this is a good practice as now with every action my app hits the recipe database (which is maybe not very elegant).
Is there a better solution to achieve this?
cache_key = "#{current_user.id}_#{unpublished_count}"
#unpublished_count = Rails.cache.fetch(cache_key, expires_in: 12.hours) do
Recipe.where(:published => false).count
end
For More: http://guides.rubyonrails.org/caching_with_rails.html#low-level-caching
To avoid hitting the database, you can introduce caching. It comes in many forms: faster storage (memcached, redis), in-process caching (global/class variables) and so on. And they all share the same problem: you need to know when to invalidate the cache.
Take a look at this guide to get some ideas: Caching with Rails.
If I were you, I would not care about this until my profiler tells me it's a performance problem. Instead, I'd direct my efforts to developing the rest of functionality.
Your falling into the trap of premature optimisation. Before doing any optimisation (which increases code complexity most of the time) you have to profile your code to find the bottleneck. Improving a SQL requests which counts for a small part of the total response time is useless. In the contrary if the SQL takes a big amount of time, that is a great improvement.
For that I can recommend these 2 gems:
https://github.com/miniProfiler/rack-mini-profiler
https://github.com/BaseSecrete/rorvswild (disclaimer: I'm the author of this one)
To reply to your question, the better way would be:
# app/models/recipe.rb
class Recipe < AR::base
# A nice scope that you can reuse anywhere
scope :unpublished, -> { where(published: false) }
end
Then in your navbar.html.erb:
<li>
<%= link_to recipes_path do %>
Recipes <span class="badge" style="background-color:#ff7373"><%= Recipe.unpublished.count %></span>
<% end %>
</li>
You have no more these ugly callback and instance variable in the controller.
Unless you have a lot of recipes (something like 100K or more) performance won't be an issue. In that case you can add an index:
CREATE INDEX index_recipes_unpblished ON recipes(published) WHERE published='f'
Note that the index applies only when published is false. Otherwise it would be counter productive.
I think that caching in your case is not well because the invalidation is extremely complex and leads to awful and easy breakable code. Don't worry to hit the database, we will never write faster code than PostgreSQL/MySQL, etc.

Ruby Find last record value in has many

I'm trying to find the last Econ_Result that belongs to a Econ_Report. I want to display the last record of the Econ_Result (ordered by "release_date") for each Econ_Report on the index view. In the controller I tried to take the list of all reports and find the last result using the following:
#econ_reports = EconReport.all
if #econ_reports.econ_results.size >= 1
#last_result = #econ_report.econ_results.last.release_date
end
econ_report.econ_results.size works on the index view when I place it in for each loop. When I try to call the value of the last record I run into issues with the fact that some reports don't yet have results (a temporary issue) so I threw in the if then check in the controller which is currently failing.
Thanks in advance for the rookie help.
Since #econ_reports is a collection of EconReport objects, you can't call an instance method like .econ_results on it. Instead, you can only call it on instances within the collection:
#econ_reports.each do |econ_report|
if econ_report.econ_results.any?
last_result = econ_report.econ_results.last
end
end
However, this can be terribly inefficient for a large collection of #econ_reports: both lines with econ_report.econ_results will query the database separately, meaning that you'll query the database independently for each econ_report in the collection. This is known as the N+1 query problem.
Luckily for you, as discussed in the link, Rails has a built-in solution to optimize this code so you'll only query the database once:
<% #econ_reports.includes(:econ_results).each do |econ_report| %>
<% if econ_report.econ_results.any? %>
<% last_result = econ_report.econ_results.last %>
# do something to display last_result
<% end %>
<% end %>
If you just want the release date you might try:
#last_result = #econ_report.econ_results.order('release_date DESC').limit(1).pluck(:release_date).first
It's worth noting that a Ruby if statement generally looks like:
if condition
end
The then is almost always omitted even though it is allowed.

Rails Russian Doll Caching and N+1

From what i understand of Russian doll caching in Rails it would be detrimental to eager load related objects or object lists when we are doing RDC (Russian Doll Caching) because in RDC we just load the top level object from the database, and look up its cached rendered template and serve. If we were to eager load related object lists that will be useless if the cache is not stale.
Is my understanding correct? If yes, how do we make sure that we eager load all the related objects on the very first call so as not to pay the cost of N+1 queries during the very first load (when the cache is not warm).
Correct - when loading a collection or a complicated object with many associations, a costly call to eager load all objects and associations can be avoided by doing a fast, simple call.
The rails guide for caching does have a good example, however, it's split up a bit. Looking at the common use case of caching a collection (ie the index action in Rails):
<% cache("products/all-#{Product.maximum(:updated_at).try(:to_i)}") do %>
All available products:
<% Product.all.each do |p| %>
<% cache(p) do %>
<%= link_to p.name, product_url(p) %>
<% end %>
<% end %>
<% end %>
This (condensed) example does 1 simple DB call Product.maximum(:updated_at) to avoid doing a much more expensive call Product.all.
For a cold cache (the second question), it is important to avoid N+1's by eager-loading associated objects. However, we know we need to do this expensive call because the first cache read for the collection missed. In Rails, this is typically done using includes. If a Product belongs to many Orders, then something like:
<% cache("products/all-#{Product.maximum(:updated_at).try(:to_i)}") do %>
All available products:
<% Product.includes(:orders).all.each do |p| %>
<% cache(p) do %>
<%= link_to p.name, product_url(p) %>
Bought at:
<ul>
<% p.orders.each do |o| %>
<li><%= o.created_at.to_s %></li>
<% end %>
</ul>
<% end %>
<% end %>
<% end %>
In the cold cache case we still do a cache read for collection and each member, however, in the partially warm cache case, we will skip rendering for a portion of the members. Note that this strategy relies on a Products associations being correctly set up to touch when associated objects are updated.
Update: This blog post describes a more complex pattern to further optimize building responses for partially cached collections. Instead of rebuilding the entire collection, it bulk fetches all available cached values then does a bulk query for the remaining values (and updates the cache). This is helpful in a couple of ways: the bulk cache read is faster than N+1 cache reads and the bulk query to the DB to build the cache is also smaller.

Minimizing number of db-queries

Let's say we have a Post model.
In the beginning of a view I need to display the total number of posts, typically by
#posts.size
Later in the view I need to display all posts, typically by
#posts.each do |post|
end
This results in two queries.
If I had done the queries in the opposite order, it had resulted in one single query (presumably utilizing CACHE).
Is there any "trick" where I can achieve the (first) mentioned order with only one db query?
You can use for example #fragment-caching
<% cache 'posts_size' do %>
<%= #posts.size %>
<% end %>
Then when a Post is created you can expire the fragment
expire_fragment 'posts_size'

Remembering records returned from DB query in Rails

I currently use a query of the form
#recordsfound = Model.where(...)
This is located in my controller and returns all records matching the query. I am using pagination to show a limited number of records at a time. However, when I select to show the next page the query is run again. Is there a way to store the records returned in a variable other than an instance variable and therefore not require the query to be rerun?
Thanks a lot guys
I'm not entirely sure I understand what you mean, but if you want the query to only run once per page even on subsequent visits, you can use fragment caching.
Rails will lazy load, and then use the cache when it hits the query in the view.
<% results.each do |result| %>
<% cache result do %>
<%= result.foo %>
<% end %>
<% end %>
If you have any dependent models, you'll have to make sure you expire the cache when they get updated if necessary:
belongs_to :result, touch: true
Note that if you are in development environment the query will still run. You can change this in your development.rb config file. If you do change this setting, don't forget to revert it . Otherwise strange things will happen and you'll waste your time trying to figure out why your changes aren't visible.
config.action_controller.perform_caching = true

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