Question

我有两种最高层阵列,其形式相同。我也想把他们合并起来:

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,产出如下:[{数值:“a”,href:“/sample1”},{数值:“b”,href:“/sample2”}>。

但是,上述法典并不奏效。仅包括第二阵列:[{数值:“b”,href:“/sample2”}]。

谁能帮助我? 提前感谢。

Answer 1

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

Answer 2

我知道两种选择:

• 构造之前的阵列,这些阵列与多种弹道:

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

或者当你把 j子组合起来时:

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,都不必打上编号。

Answer 3

尤里的回答让我结束,但我的最终解决办法是,在我的<代码>.jbuilder档案中这样做。

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

我把阵列排列的顺序是组合阵列的顺序。

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