Question

我有两个顶级数组，它们具有相同的格式。我想合并它们：

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

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

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

有人可以帮助我吗？提前谢谢。

Answer 1

我知道两个选项：

在迭代之前组合数组，这适用于多个鸭子源数组：

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

或者当你有鸭子和火鸡时，结合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，都不需要调用#to_json或类似的东西。

Answer 2

Yuri的回答让我很接近，但对我而言，最终的解决方案就是在我的.jbuilder文件中执行此操作。

json.array!(companies) do |company|
  json.value(company.to_s)
  json.href(employee_company_path(company))
end

json.array!(company_people) do |person|
  json.value(person.to_s)
  json.href(employee_company_path(person.company))
end

我放入数组的顺序是数组组合的顺序。

Answer 3

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

Jbuilder：我如何合并2个顶级数组？

3 个答案: