How to access complex array of hashes

Question

I have this to work with:

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

I want to be able to access each name and path so I tried:

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

but this returns an Enumerator. Any idea how I can do this?

Consider creating your own Class for that kind of thing - more readable. — Reactormonk
– Reactormonk, Commented Mar 22, 2012 at 15:35

Ivaylo Strandjev · Accepted Answer · 2012-03-22 15:34:03Z

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

This will handle even a bit more complex cases when you have different structure for different elements.

answered Mar 22, 2012 at 15:34

Ivaylo Strandjev

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1 Comment

Josh Over a year ago

This is what I ended up doing- apps = apps.flatten apps = apps.delete_if do |key, value| key.class == String end apps.each do |app| puts blah end

kclair · Accepted Answer · 2012-03-22 15:30:17Z

0

I think your first each loop is only looping over an array, so it would be:

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

answered Mar 22, 2012 at 15:30

kclair

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Comments

steenslag · Accepted Answer · 2012-03-22 20:40:33Z

0

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"]}

answered Mar 22, 2012 at 20:40

steenslag

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