我正在使用python2.7
我有一个json我拉，当我请求时它总是在变化。
我需要在我的词典中的Relation6下的Term7下提取Animal_Target_DisplayName。
问题是有时Relation6对象位于Json的另一部分中，它可能会更深层次或以其他顺序进行调平。
我正在尝试创建仅可以导出键Animal_Target_DisplayName的值的代码，但是没有任何效果。它甚至不会循环嵌套的字典。
现在，如果我只是使用['view']['Term0'][0]['Relation6']之类的东西将其拔出，但是请记住，JSON永远不会以相同的结构返回。
我正在使用代码获取键Animal_Target_DisplayName的值，但是它似乎并没有遍历我的字典并找到具有该键的所有值。
array = []
for d in dict.values():
row = d['Animal_Target_DisplayName']
array.append(row)
下方的JSON：
dict = {
"view":{
"Term0":[
{
"Id":"b0987b91-af12-4fe3-a56f-152ac7a4d84d",
"DisplayName":"Dog",
"FullName":"Dog",
"AssetType1":[
{
"AssetType_Id":"00000000-0000-0000-0000-000000031131",
}
]
},
{
"Id":"ee74a59d-fb74-4052-97ba-9752154f015d",
"DisplayName":"Dog2",
"FullName":"Dog",
"AssetType1":[
{
"AssetType_Id":"00000000-0000-0000-0000-000000031131",
}
]
},
{
"Id":"eb548eae-da6f-41e8-80ea-7e9984f56af6",
"DisplayName":"Dog3",
"FullName":"Dog3",
"AssetType1":[
{
"AssetType_Id":"00000000-0000-0000-0000-000000031131",
}
]
},
{
"Id":"cfac6dd4-0efa-4417-a2bf-0333204f8a42",
"DisplayName":"Animal Set",
"FullName":"Animal Set",
"AssetType1":[
{
"AssetType_Id":"00000000-0000-0000-0001-000400000001",
}
],
"StringAttribute2":[
{
"StringAttribute_00000000-0000-0000-0000-000000003114_Id":"00a701a8-be4c-4b76-a6e5-3b0a4085bcc8",
"StringAttribute_00000000-0000-0000-0000-000000003114_Value":"Desc"
}
],
"StringAttribute3":[
{
"StringAttribute_00000000-0000-0000-0000-000000000262_Id":"a81adfb4-7528-4673-8c95-953888f3b43a",
"StringAttribute_00000000-0000-0000-0000-000000000262_Value":"meow"
}
],
"BooleanAttribute4":[
{
"BooleanAttribute_00000000-0000-0000-0001-000500000001_Id":"932c5f97-c03f-4a1a-a0c5-a518f5edef5e",
"BooleanAttribute_00000000-0000-0000-0001-000500000001_Value":"true"
}
],
"SingleValueListAttribute5":[
{
"SingleValueListAttribute_00000000-0000-0000-0001-000500000031_Id":"ef51dedd-6f25-4408-99a6-5a6cfa13e198",
"SingleValueListAttribute_00000000-0000-0000-0001-000500000031_Value":"Blah"
}
],
"Relation6":[
{
"Animal_Id":"2715ca09-3ced-4b74-a418-cef4a95dddf1",
"Term7":[
{
"Animal_Target_Id":"88fd0090-4ea8-4ae6-b7f0-1b13e5cf3d74",
"Animal_Target_DisplayName":"Animaltheater",
"Animal_Target_FullName":"Animaltheater"
}
]
},
{
"Animal_Id":"6068fe78-fc8e-4542-9aee-7b4b68760dcd",
"Term7":[
{
"Animal_Target_Id":"4e87a614-2a8b-46c0-90f3-8a0cf9bda66c",
"Animal_Target_DisplayName":"Animaltitle",
"Animal_Target_FullName":"Animaltitle"
}
]
},
{
"Animal_Id":"754ec0e6-19b6-4b6b-8ba1-573393268257",
"Term7":[
{
"Animal_Target_Id":"a8986ed5-3ec8-44f3-954c-71cacb280ace",
"Animal_Target_DisplayName":"Animalcustomer",
"Animal_Target_FullName":"Animalcustomer"
}
]
},
{
"Animal_Id":"86b3ffd1-4d54-4a98-b25b-369060651bd6",
"Term7":[
{
"Animal_Target_Id":"89d02067-ebe8-4b87-9a1f-a6a0bdd40ec4",
"Animal_Target_DisplayName":"Animalfact_transaction",
"Animal_Target_FullName":"Animalfact_transaction"
}
]
},
{
"Animal_Id":"ea2e1b76-f8bc-46d9-8ebc-44ffdd60f213",
"Term7":[
{
"Animal_Target_Id":"e398cd32-1e73-46bd-8b8f-d039986d6de0",
"Animal_Target_DisplayName":"Animalfact_transaction",
"Animal_Target_FullName":"Animalfact_transaction"
}
]
}
],
"Relation10":[
{
"TargetRelation_b8b178ff-e957-47db-a4e7-6e5b789d6f03_Id":"aff80bd0-a282-4cf5-bdcc-2bad35ddec1d",
"Term11":[
{
"AnimalId":"3ac22167-eb91-469a-9d94-315aa301f55a",
"AnimalDisplayName":"Animal",
"AnimalFullName":"Animal"
}
]
}
],
"Tag12":[
{
"Tag_Id":"75968ea6-4c9f-43c9-80f7-dfc41b24ec8f",
"Tag_Name":"AnimalAnimaltitle"
},
{
"Tag_Id":"b1adbc00-aeef-415b-82b6-a3159145c60d",
"Tag_Name":"Animal2"
},
{
"Tag_Id":"5f78e4dc-2b37-41e0-a0d3-cec773af2397",
"Tag_Name":"AnimalDisplayName"
}
]
}
]
}
}
我要获取的输出是键Animal_Target_DisplayName的所有值的列表，例如['Animaltheater','Animaltitle', 'Animalcustomer', 'Animalfact_transaction', 'Animalfact_transaction']，但我们需要记住，此json的嵌套结构始终在变化，但其键是总是一样。
答案 0 :(得分：1)
我想您唯一的选择正在整个字典中运行并获取Animal_Target_DisplayName键的值，我提出了以下递归解决方案：
def run_json(dict_):
animal_target_sons = []
if type(dict_) is list:
for element in dict_:
animal_target_sons.append(run_json(element))
elif type(dict_) is dict:
for key in dict_:
if key=="Animal_Target_DisplayName":
animal_target_sons.append([dict_[key]])
else:
animal_target_sons.append(run_json(dict_[key]))
return [x for sublist in animal_target_sons for x in sublist]
run_json(dict_)
然后调用run_json将返回带有所需内容的列表。顺便说一句，我建议您将json从dict重命名为dict_，因为dict是Python对于字典类型的保留字。
答案 1 :(得分：-1)
既然获取JSON，为什么不使用json模块呢？这样将为您进行解析，并允许您使用字典功能+功能来获取所需的信息。
#!/usr/bin/python2.7
from __future__ import print_function
import json
# _somehow_ get your JSON in as a string. I'm calling it "jstr" for this
# example.
# Use the module to parse it
jdict = json.loads(jstr)
# our dict has keys...
# view -> Term0 -> keys-we're-interested-in
templist = jdict["view"]["Term0"]
results = {}
for _el in range(len(templist)):
if templist[_el]["FullName"] == "Animal Set":
# this is the one we're interested in - and it's another list
moretemp = templist[_el]["Relation6"]
for _k in range(len(moretemp)):
term7 = moretemp[_k]["Term7"][0]
displayName = term7["Animal_Target_DisplayName"]
fullName = term7["Animal_Target_FullName"]
results[fullName] = displayName
print("{0}".format(results))
然后，您可以将results字典简单地或漂亮地打印出来：
>>> print(json.dumps(results, indent=4))
{
"Animaltitle2": "Animaltitle2",
"Animalcustomer3": "Animalcustomer3",
"Animalfact_transaction4": "Animalfact_transaction4",
"Animaltheater1": "Animaltheater1"
}