Python：替换整个json中嵌套的dict键的值，无论是在列表中还是直接在dict中

Question

我想将json中的值替换为dictionary的键，它可以直接在json中显示为dictionary，也可以在另一个列表中显示为dictionary，如下所示：

{ 
  "appType": "popper",
  "createdAt": "1970-01-01T00:00:00.000Z",
  "updatedAt": "1970-01-01T00:00:00.000Z",
  "people": [{
            "name": "Vol1",
            "label": "Vol1",
            "peopleInfo": [{
                "name": "ram",
                "age": "2407653459860",
                "id": "1b738651-da9f-4c85-88c1-70dbfe1976681"
            }],
            "itemInfo": {
                "id": "ee763970-51e2-57a5-955c-d72fc3e28a3f",
                "name": "xyz",
                "type": "any",
                "managed": False
            }
        }],
  "itemInfo": [{
            "managed": False,
            "vendorName": "any",
            "serialNumber": "AF-10124"
             }],
}

期望输出：

{ 
  "appType": "popper",
  "createdAt": "1970-01-01T00:00:00.000Z",
  "updatedAt": "1970-01-01T00:00:00.000Z",
  "peopleInfo": [{
            "name": "Vol1",
            "label": "Vol1",
            "people": [{
                "name": "ram",
                "age": "2407653459860",
                "id": "1b738651-da9f-4c85-88c1-70dbfe1976681"
            }],
            "itemInfo": {
                "id": "ee763970-51e2-57a5-955c-d72fc3e28a3f",
                "name": "xyz",
                "type": "any",
                "managed": True
            }
        }],
  "itemInfo": [{
            "managed": True,
            "vendorName": "any",
            "serialNumber": "AF-10124"
             }],
}

因此，正如在所需的输出中一样，我希望在json中直接为itemInfo更新/替换managed标记为True，以及使用python在peopleInfo列表中更新/替换itemInfo。iteminfo字典也可以存在于一些不同列表中的整个json中。谢谢你的帮助

我已经编写了以下代码，但无法使其成为通用代码：

i["details"]["storageSystemsInfo"][0]["managed"] = True

Answer 1

更新2：

按照您的要求，仅更改“itemInfo”字典中的“托管”字段。 （这不是漂亮或干净的代码）

# Python 3
def change_this_key_to_true(key, var):
    if hasattr(var,'items'):
        for k, v in var.items(): 
            if k == key:
                var[k] = True
            if isinstance(v, dict):
                change_this_key_to_true(key, v)
            elif isinstance(v, list):
                for d in v:
                    change_this_key_to_true(key, d)
    else:
        if isinstance(var, list):
            for element in var:
                change_this_key_to_true(key, element)


all_item_info_dics = []
def get_all_values_with_key(key, var):
    if hasattr(var, 'items'):
        for k, v in var.items(): 
            if k == key:
                all_item_info_dics.append(var[k])
            if isinstance(v, dict):
                get_all_values_with_key(key, v)
            elif isinstance(v, list):
                for d in v:
                    get_all_values_with_key(key, d)

get_all_values_with_key("itemInfo", myJson)
print(all_item_info_dics)
print("\n")
change_this_key_to_true("managed", all_item_info_dics)
print(myJson)

更新： （这是我的第一个答案，花了我一段时间，希望是好的：））

# Python 3
def change_this_key_to_true(key, var):
    if hasattr(var,'items'):
        for k, v in var.items():
            if k == key:
                var[k] = True
            if isinstance(v, dict):
                change_this_key_to_true(key, v)
            elif isinstance(v, list):
                for d in v:
                    change_this_key_to_true(key, d)

change_this_key_to_true("managed", myJson)
print(myJson)

在复杂的dict/list/“这两者的奇怪组合”中遍历所有字段，查找名为“managed”的字段并将其更改为True。 根据我的回答： Find all occurrences of a key in nested dictionaries and lists

原始答复：

x["people"][0]["itemInfo"]["managed"] = True
x["itemInfo"][0]["managed"] = True