Question

我尝试返回name my.name为{ "total_results"=>"73", "results"=> { "28"=> {"description"=>"description_here", "name"=> "my.name"}, "25"=> {"description"=>"description_here", "name"=> "other1.name"}, "24"=> {"description"=>"description_here", "name"=> "my.name"}, "21"=> {"description"=>"description_here", "name"=> "other2.name"} } }的最低结果数字：

因此，在上面的示例中，它将返回Process: com.example.nativeaudio, PID: 5895 java.lang.UnsatisfiedLinkError: dalvik.system.PathClassLoader[DexPathList[[zip file "/data/app/com.example.nativeaudio-2/base.apk"],nativeLibraryDirectories=[/data/app/com.example.nativeaudio-2/lib/x86, /vendor/lib, /system/lib]]] couldn't find "libnative-audio-jni.so" at java.lang.Runtime.loadLibrary(Runtime.java:367) at java.lang.System.loadLibrary(System.java:1076) at com.example.nativeaudio.NativeAudio.<clinit>(NativeAudio.java:313) at java.lang.Class.newInstance(Native Method) at android.app.Instrumentation.newActivity(Instrumentation.java:1067) at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2317) at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2476) at android.app.ActivityThread.-wrap11(ActivityThread.java) at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1344) at android.os.Handler.dispatchMessage(Handler.java:102) at android.os.Looper.loop(Looper.java:148) at android.app.ActivityThread.main(ActivityThread.java:5417) at java.lang.reflect.Method.invoke(Native Method) at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:726) at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:616)。我对Ruby很新，所以我很难找到一个很好的方法来做到这一点。

Answer 1

你的例子是Ruby中的哈希

h = {
"total_results"=>"73", 
"results"=>
    {
    "28"=> {"description"=>"description_here", "name"=> "my.name"},
    "25"=> {"description"=>"description_here", "name"=> "other1.name"},
    "24"=> {"description"=>"description_here", "name"=> "my.name"},
    "21"=> {"description"=>"description_here", "name"=> "other2.name"}
    }
}

您可以使用select获取所有匹配的值，然后获取它们的min键，如

h["results"].select{|k,v| v["name"] == "my.name"}.keys.map(&:to_i).min
# => 24

Answer 2

这是一个解决方案。绝对不是最好的，但我至少可以告诉你如何遍历那个哈希..

foo = {
"total_results"=>"73", 
"results"=>
    {
    "28"=> {"description"=>"description_here", "name"=> "my.name"},
    "25"=> {"description"=>"description_here", "name"=> "other1.name"},
    "24"=> {"description"=>"description_here", "name"=> "my.name"},
    "21"=> {"description"=>"description_here", "name"=> "other2.name"}
    }
}

bar = nil

foo['results'].each do |k, v|
  if v['name'] == "my.name"
    if bar.nil? || k < bar
      bar = k
    end
  end
end


puts bar  #=> "24"

Answer 3

试试这个：

h = {
    "total_results"=>"73", 
    "results"=>
        {
        "28"=> {"description"=>"description_here", "name"=> "my.name"},
        "25"=> {"description"=>"description_here", "name"=> "other1.name"},
        "24"=> {"description"=>"description_here", "name"=> "my.name"},
        "21"=> {"description"=>"description_here", "name"=> "other2.name"}
        }
}

获取最小密钥：

min_key = h["results"].keys.min

获取 min_key 值

h["results"][min_key]

Answer 4

单线解决方案

鉴于您在 h 中存储了已发布的哈希值，您可以执行以下操作：

h['results'].select { |k,v| v['name'] == 'my.name' }.keys.min_by(&:to_i)
#=> "24"

说明

虽然为了清晰起见，这个单行代表交换紧凑性，但它正在做以下事情：

使用Enumerable#select筛选带有嵌套哈希键“name”的值的结果。

返回包含哈希值的临时哈希值，其中分配给“name”键的值包含所需的字符串值。例如：

{"28"=>{"description"=>"description_here", "name"=>"my.name"},
 "24"=>{"description"=>"description_here", "name"=>"my.name"}}

使用Enumerable#min_by强制键进行比较，然后找到最小值并以原始格式返回（例如作为字符串）。

只有看起来就像魔法一样。这只是Ruby。：）

Answer 5

h = {
    "total_results"=>"73", 
    "results"=>
        {
        "28"=> {"description"=>"description_here", "name"=> "my.name"},
        "25"=> {"description"=>"description_here", "name"=> "other1.name"},
        "24"=> {"description"=>"description_here", "name"=> "my.name"},
        "21"=> {"description"=>"description_here", "name"=> "other2.name"}
        }
} 

h["results"].select{|key,hash| [key, hash] if hash["name"]=="my.name"}.keys.map(&:to_i).min

=> 24

Answer 6

仅通过一行代码查找min_key：

min_key = your_hash['results'].sort.to_h.invert['my.name']

如何在Ruby中解析和查找JSON元素？

6 个答案:

单线解决方案

说明