如何遍历JSON嵌套对象？解决'dict' object is not callable报错

问题重现 #

你处理的JSON片段：

{"id":"2769","data":{"accommodation":{"town":"Dublin","addressYears":1,"postcode":"K12 Y453","accommodationType":"select"},"bankDetails":{"payee":"Name","Number":"1234567890"},"creditReport":{"creditReportMessageId":"message id","creditReportContractCode":"contact code","creditReportSource":"cccs"},"dependents":{"dependentsCount":0,"dependentsCount2":0}}}

你尝试的遍历代码：

for i in data:
    print(i)
    for j in data[i]:
        print(j)
        for k in data(j):
            print(k)

运行时出现错误：

'dict' object is not callable

错误原因 #

第三层循环里的data(j)是错误写法：Python中字典的元素访问必须用方括号[]，而括号()是用来调用函数的。这里data是字典对象，不是可调用的函数，因此触发该错误。此外，你要遍历的是第二层字典（比如data[i]对应的accommodation字典）的子键，第三层应该遍历data[i][j]的键。

修正后的代码 #

假设你已经将JSON字符串解析为Python字典（用json.loads()），修正后的遍历代码如下：

import json

# 解析JSON字符串
json_content = '{"id":"2769","data":{"accommodation":{"town":"Dublin","addressYears":1,"postcode":"K12 Y453","accommodationType":"select"},"bankDetails":{"payee":"Name","Number":"1234567890"},"creditReport":{"creditReportMessageId":"message id","creditReportContractCode":"contact code","creditReportSource":"cccs"},"dependents":{"dependentsCount":0,"dependentsCount2":0}}}'
parsed_data = json.loads(json_content)
data = parsed_data['data']

# 修正后的三层循环
for first_key in data:
    print(first_key)
    second_dict = data[first_key]
    for second_key in second_dict:
        print(f"  {second_key}")
        third_dict = second_dict[second_key]
        for third_key in third_dict:
            print(f"    {third_key}")

运行后会按层级输出所有键：

accommodation
  town
    (无第三层，对应值为字符串)
  addressYears
    ...
  postcode
    ...
  accommodationType
    ...
bankDetails
  payee
    ...
  Number
    ...
...

更通用的递归遍历方案 #

如果JSON的嵌套层级不确定，递归方法能更灵活地处理所有层级的键：

def traverse_nested_keys(dictionary, indent=0):
    for key in dictionary:
        print("  " * indent + key)
        value = dictionary[key]
        # 如果当前值还是字典，继续递归遍历
        if isinstance(value, dict):
            traverse_nested_keys(value, indent + 1)

# 调用函数遍历所有层级键
traverse_nested_keys(data)

这个方法会自动识别嵌套字典，逐层打印所有键，无需手动编写多层循环。

内容的提问来源于stack exchange，提问作者Jordan121