Tim*_* T. 1 controller ruby-on-rails dry
我有一个控制器有两个不同的动作,但两个都需要相同的代码,这有点长,我怎么能让他们访问相同的行为,但保持干燥?
@list = Contact.find :all,
:select => "companies.name AS co_name,
companies.id AS comp_id,
COUNT(contact_emails.id) AS email_count,
COUNT(contact_calls.id) AS call_count,
COUNT(contact_letters.id) AS letter_count,
COUNT(contact_postalcards.id) AS postalcard_count",
:conditions => ['contact_emails.date_sent < ? and contact_emails.date_sent > ?',
report_end_date, report_start_date],
:joins => [
"LEFT JOIN companies ON companies.id = contacts.company_id",
"LEFT JOIN contact_emails ON contact_emails.contact_id = contacts.id",
"LEFT JOIN contact_letters ON contact_letters.contact_id = contacts.id",
"LEFT JOIN contact_postalcards ON contact_postalcards.contact_id = contacts.id",
"LEFT JOIN contact_calls ON contact_calls.contact_id = contacts.id"
],
#:group => "companies.id"
:group => "companies.name"
puts @list[0].attributes.inspect
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您应该将此代码移动到模型:
# Contatct model
def self.get_list(report_start_date, report_end_date)
self.find :all,
:select => "companies.name AS co_name,
companies.id AS comp_id,
COUNT(contact_emails.id) AS email_count,
COUNT(contact_calls.id) AS call_count,
COUNT(contact_letters.id) AS letter_count,
COUNT(contact_postalcards.id) AS postalcard_count",
:conditions => ['contact_emails.date_sent < ? and contact_emails.date_sent > ?',
report_end_date, report_start_date],
:joins => [
"LEFT JOIN companies ON companies.id = contacts.company_id",
"LEFT JOIN contact_emails ON contact_emails.contact_id = contacts.id",
"LEFT JOIN contact_letters ON contact_letters.contact_id = contacts.id",
"LEFT JOIN contact_postalcards ON contact_postalcards.contact_id = contacts.id",
"LEFT JOIN contact_calls ON contact_calls.contact_id = contacts.id"
],
#:group => "companies.id"
:group => "companies.name"
end
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然后你可以在控制器中使用它:
@list = Contact.get_list(report_start_date, report_end_date)
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也许您也可以将它拆分为较小的部分,并使用范围和已定义的关联,而不是自己编写所有关联.