我可以根據虛擬屬性來計算嗎？

Question

我有以下錯誤：

no such column: company_name: SELECT count("contact_emails".id) AS count_id FROM "contact_emails" 

模型ContactEmail沒有列COMPANY_NAME。我將它創建爲一個虛擬屬性。

不可能根據這些信息做選擇嗎？那麼我該怎麼做呢？

ContactEmail belongs_to聯繫人belongs_to公司。

Answer 1

添加到類定義但在數據庫中不存在的虛擬屬性不能用於數據庫選擇，因爲虛擬屬性不在數據庫中。

您可以使用數據庫選擇所需行的超集，然後在將使用虛擬屬性的Rails級別上進行第二次選擇。

如

# model Citizen class file 
# model fields: 
# id 
# name 
# age 
# city 

def can_vote? # a virtual attribute 
    age >= 18 
end 

def self.find_voters_by_city(city) # class level finder 
    # returns array of voters in a city 
    citizens = Citizen.find_by_city(city) # First select done in database 
    citizens.select{|citizen| citizen.can_vote?} # Second select is done at Rails 
               # level using the Array#select 
               # method 
end 

注意的是，雖然上述工作正常，你應該非常小心，在性能方面的問題。在Rails級別選擇比在dbms中選擇要慢得多。此外，您通過Rails/DBMS連接傳輸的數據遠多於其他方式需要的數據。

如果您打算定期選擇某些東西，那麼將虛擬屬性推入數據庫通常會更好 - 使其成爲數據庫中的真正屬性。

如果您的表格無法更改，您還可以使用has_one關係創建第二個表格。第二個表格將保存附加屬性。

新增：使用兩個表

### model Contact 
# id 
# name 
# city 
has_one :company 
def company_name; company.name; end # a virtual attribute 

### model Company 
# id 
# contact_id 
# name 
# employee_count 
belongs_to :contact 

### Selecting on city (from Contact) and employee_count (from Company) 
city = xyz # the city we are searching for 
employee_count = 123 # minimum company size we want 
contacts = Contact.find(:all, 
     :conditions => ["contacts.city = ? and companies.employee_count >= ?", 
     city, employee_count], 
     :include => :company) 

# above statement does the join and select in the database 
# There are also techniques involving named scopes for the above 
# And in Rails 3, it will be different too.