Jbuilder：如何合并两个顶级数组？

Question

我有两个顶级数组，它们具有相同的格式。我想将它们合并：

json = Jbuilder.encode do |json|
  json.(companies) do |json, c|
    json.value c.to_s
    json.href employee_company_path(c)
  end
  json.(company_people) do |json, cp|
    json.value "#{cp.to_s} (#{cp.company.to_s})"
    json.href employee_company_path(cp.company)
  end
end

因此输出如下： "[{value: "a", href: "/sample1"}, {value: "b", href: "/sample2"}]"

但是上面的代码不起作用。它仅包含第二个数组："[{value: "b", href: "/sample2"}]"

有人可以帮我吗？提前致谢。

Answer 1

我知道两个选择：

1. 在迭代之前合并数组，这与鸭子的多个源数组很好地配合：

   def Employee
     def company_path
       self.company.company_path if self.company
     end
   end
   
   [...]
   
   combined = (companies + company_people).sort_by{ |c| c.value }
   # Do other things with combined
   
   json.array!(combined) do |duck|
     json.value(duck.to_s)
     json.href(duck.company_path)
   end
   

2. 或者，当您有鸭子和火鸡时，请组合json数组：

   company_json = json.array!(companies) do |company|
     json.value(company.to_s)
     json.href(employee_company_path(company))
   end
   
   people_json = json.array!(company_people) do |person|
     json.value(person.to_s)
     json.href(employee_company_path(person.company))
   end
   
   company_json + people_json
   

在这两种情况下，都无需调用#to_json或类似名称。

Answer 2

result =  []
companies.each do |c|
  result << {:value => c.to_s, :href => employee_company_path(c)
end
company_people.each do |c|
  result << {:value => "#{cp.to_s} (#{cp.company.to_s})", :href => employee_company_path(cp.company)
end
# at this point result will be an array of companies and people which just needs converting to json.
result.to_json