Question

我希望能够使用单个“ nested_key”变量从一组（任意深度）的嵌套字典中获取一个值。

如果我有字典，可以像这样简单地将其编入索引：

create procedure readDataFromJSON @data nvarchar(4000)
as
begin

  select 
   contributor,difficulty,rating,instructions,[name],preptime,cooktime,imgurl,protien,fats,carbs,fiber,reciepdescription
   into #tmprecipe
   from OPENJSON(@data,'$.Rootobject.recipe')
   WITH (
      RecipeContributor varchar(100) '$.RecipeContributor' as contributor,  
      RecipeDifficulty varchar(100) '$.RecipeDifficulty' as difficulty, 
      RecipeRating varchar(100) '$.RecipeRating' as rating, 
      RecipeInstructions varchar(100) '$.RecipeInstructions' as instructions, 
      RecipeName varchar(100) '$.RecipeName' as name, 
      RecipePrepTime varchar(100) '$.RecipePrepTime' as preptime, 
      RecipeCookTime varchar(100) '$.RecipeCookTime' as cooktime, 
      ImageURL varchar(100) '$.ImageURL' as imgurl, 
      RecipeProtein varchar(100) '$.RecipeProtein' as protien, 
      RecipeFats varchar(100) '$.RecipeFats' as fats, 
      RecipeCarbs varchar(100) '$.RecipeCarbs' as carbs, 
      RecipeFiber varchar(100) '$.RecipeFiber' as fibre, 
      RecipeDescription varchar(100) '$.RecipeDescription' as reciepdescription, 
   );

   select 
     quantity,sizename,ingredientname
     into #tmprecipeingrident
     from OPENJSON(@data,'$.Rootobject.recipeIngredients')
     WITH (
             IngredientListQuanity varchar(100) '$.IngredientListQuanity' as quantity,  
             MeasurementSizeName varchar(100) '$.MeasurementSizeName'  as sizename,
             IngredientName varchar(100) '$.IngredientName'  as ingredientname
     );

     select * from #tmprecipe;
     select * from #tmprecipeingrident;

     drop table #tmprecipeingrident;
     drop table #tmprecipe;
end

但是我不认为（？）我可以轻松地将索引链变成变量并以相同的方式使用它？

a = {"Alice": {"Car": {"Color": "Blue"}}}
a["Alice"]["Car"]["Color"]
>>> "Blue"

列表具有a = {"Alice": {"Car": {"Color": Blue}}} nested_key = ["Alice"]["Car"]["Color"] # for use in: a[nested_key] >>> TypeError: list indicies must be integer, not str对象；但是有没有办法为变量分配“类似键”的对象？

否则，我认为最好的选择是slice()样式函数（例如https://stackoverflow.com/a/28225660/2588039）？

Answer 1

您可以创建一个在键上循环的函数，每次在字典上索引索引项，并带有给定的键列表：

def get_nested_value(dct, key_list)
    value = dct 
    for key in key_list:
        value = value[key]

    return value 

keys = ['Alice', 'Car', 'Colour']
print(get_nested_value(a, keys))

带变量的字典的索引字典

1 个答案: