Question

我可以使用它：

[
  ["app1", {"name"=>"name1", "path"=>"xyz.com/"}],
  ["app2", {"name"=>"name2", "path"=>"xyz.com/"}],
  ["app3", {"name"=>"name3", "path"=>"xyz.com/"}],
  # etc.
]

我希望能够访问每个名称和路径，所以我尝试了：

apps.each do |key, value|
  value.each do |key, value|
    puts value
  end
end

但这会返回一个枚举器。知道我怎么能这样做吗？

Answer 1

apps = [["app1", {"name"=>"name1", "path"=>"https://xyz.com/"}], ["app2", {"name"=>"name2", "path"=>"https:/xyz.com/"}], ["app3", {"name"=>"name3", "path"=>"https://xyz.com/"}]]
apps.flatten.each do |t|
  next unless t.class == Hash
  next unless t.key?("name")
  next unless t.key?("path")
  puts t.inspect # now t is a hash that has both "name" and "path" keys - do what you want
end

当你有不同元素的不同结构时，这将处理更复杂的情况。

Answer 2

我认为你的第一个循环只是循环遍历一个数组，所以它将是：

apps.each do |app|
  app.each do |key, value|
    puts key   # would be app1 in the first array
    puts value["name"]
    puts value["path"]
  end
end

Answer 3

ar = [
  ["app1", {"name"=>"name1", "path"=>"xyz.com/"}],
  ["app2", {"name"=>"name2", "path"=>"xyz.com/"}],
  ["app3", {"name"=>"name3", "path"=>"xyz.com/"}]

]
#Get a specific app:
p ar.assoc("app2").last["name"]
#Get all names and paths
ar.each{|app|  name, path = app.last["name"], app.last["path"]}

如何访问复杂的哈希数组

3 个答案: