Wes*_*ter 3 ruby-on-rails ruby-on-rails-4
我们的目标是选择Store哪一个中Coupon使用最多的。
目前,我有这个,它有效(分解以供解释):
# coupon.rb
has_many :redemptions
has_and_belongs_to_many :stores
def most_popular_store
stores.find( # Return a store object
redemptions # Start with all of the coupon's redemptions
.group(:store_id) # Group them by the store_id
.count # Get a hash of { 'store_id' => 'count' } values
.keys # Create an array of keys
.sort # Sort the keys so highest is first
.first # Take the ID of the first key
)
end
###
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它是这样使用的:
describe 'most_popular_store' do
it 'returns the most popular store' do
# Create coupon
coupon = FactoryGirl.create(:coupon)
# Create two different stores
most_popular_store = FactoryGirl.create(:store, coupons: [coupon])
other_store = FactoryGirl.create(:store, coupons: [coupon])
# Add redemptions between those stores
FactoryGirl.create_list(:redemption, 2, coupon: coupon, store: other_store)
FactoryGirl.create_list(:redemption, 5, coupon: coupon, store: most_popular_store)
# Verify
expect(coupon.most_popular_store.title).to eq most_popular_store.title
end
end
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就像我说的,该方法有效,但它看起来像猴子补丁。如何重构我的most_popular_store方法?
小智 6
我认为你的方法实际上不起作用。count给你一个键作为 store_ids 和值作为计数keys的散列,然后你在散列上运行,它给你一个 store_ids 数组。从那时起,您已经失去了计数,您正在按 store_ids 排序并获取第一个。您的测试通过的唯一原因是您在另一个之前创建了流行商店,因此它的 id 较低(sort默认情况下按升序排序)。要获得正确的结果,请进行以下更改:
redemptions # Start with all of the coupon's redemptions
.group(:store_id) # Group them by the store_id
.count # Get a hash of { 'store_id' => 'count' } values
.max_by{|k,v| v} # Get key, val pair with the highest value
# output => [key, value]
.first # Get the first item in array (the key)
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