Question

鉴于此数组：

[[{"RepCode"=>"AL20", "ID"=>"eae71dff-3796-4c61-956e-a75a00b01a7b", "Name"=>"Schuh, Eddy", "Folios"=>[]}],
[{"RepCode"=>"ABNX", "ID"=>"637e9117-ed03-45ef-8950-a7220087ee9a", "Name"=>"Eckerson, Kathy", "Folios" => [{"ID"=>"d0cda2be-c142-47d1-9a81-a76c0eea2765"}],
[{"RepCode"=>"ABCD", "ID"=>"637e9117-ed03-45ef-8950-a234902038", "Name"=>"Sarah, Barber", "Folios" => [{"ID"=>"46aafe31-f686-49e2-9d58-c694ea55c14f"}]]

我需要返回与<{1}}键的给定ID匹配的 ONE 数组

CODE：

Folio

返回：

correct_manager = managers.detect do |manager|
  manager.first["Folios"].map { |f| f["ID"] == "d0cda2be-c142-47d1-9a81-a76c0eea2765" }
end

我希望它返回

{"RepCode"=>"AL20", "ID"=>"eae71dff-3796-4c61-956e-a75a00b01a7b", "Name"=>"Schuh, Eddy", "Folios"=>[]}

因为ID在检测方法中匹配。

如何返回与传入的id匹配的一个数组？

Answer 1

您可以使用Enumerable #find

correct_manager = managers.find do |manager|
  folios = manager.first["Folios"][0] || {}
  folios["ID"] == "d0cda2be-c142-47d1-9a81-a76c0eea2765"
end

Answer 2

def doit(managers, val)
  managers.find { |(h)| h["Folios"] == ["ID"=>val] }
end

managers = [
  [{ "RepCode"=>"AL20", "Folios"=>[] }],
  [{ "RepCode"=>"ABNX", "Folios"=>[{ "ID"=>"d0cda2be-c142-47d1-9a81" }] }],
  [{ "RepCode"=>"ABCD", "Folios"=>[{ "ID"=>"46aafe31-f686-49e2-9d58" }] }]
]

doit(managers, "d0cda2be-c142-47d1-9a81")
  # => [{"RepCode"=>"ABNX", "Folios"=>[{"ID"=>"d0cda2be-c142-47d1-9a81"}]}]

Answer 3

首先让我们正确格式化您的数据：

managers =  [
  [ {"RepCode"=>"AL20", "ID"=>"eae71dff-3796-4c61-956e-a75a00b01a7b", "Name"=>"Schuh, Eddy", "Folios"=>[] } ],
  [ {"RepCode"=>"ABNX", "ID"=>"637e9117-ed03-45ef-8950-a7220087ee9a", "Name"=>"Eckerson, Kathy", "Folios" => [{"ID"=>"d0cda2be-c142-47d1-9a81-a76c0eea2765"}] } ],
  [ {"RepCode"=>"ABCD", "ID"=>"637e9117-ed03-45ef-8950-a234902038", "Name"=>"Sarah, Barber", "Folios" => [{"ID"=>"46aafe31-f686-49e2-9d58-c694ea55c14f"}] } ] 
]

target_id = 'd0cda2be-c142-47d1-9a81-a76c0eea2765'

managers.flatten.find{|k,_v| k['Folios'].any?{|f| f.key?('ID') && f['ID'] == target_id}}

如果满足条件，则返回数组中的元素

CODE：

3 个答案: