Python查找不限层级Json数据中某个key或者value的路径方式

最近项目中有一个小需求，查找json文件中某个key或者value的路径，所以就写了一个简单的小脚本，比较粗糙。

#!/usr/bin/env python3
# -*- coding:utf-8 -*-
@author: funcups
from logzero import logger
import ast
class HandleJson():
  def __init__(self, data):
    if data == None:
      logger.error('请输入json格式数据')
      exit()
    if isinstance(data, str):
        self.data = ast.literal_eval(data)
      except:
        logger.error('请输入正确的json格式数据')
        exit()
    elif isinstance(data, dict):
      self.data = data
  def __paths(self, data, path=''):
    用于遍历json树
    :param data: 原始数据，或者key对应的value值
    :param path: key值字符串，默认值为''
    :return:
    if isinstance(data, dict):
      for k, v in data.items():
        tmp = path + "['%s']" % k
        yield (tmp, v)
        yield from self.__paths(v, tmp)
    if isinstance(data, list):
      for k, v in enumerate(data):
        tmp = path + '[%d]' % k
        yield (tmp, v)
        yield from self.__paths(v, tmp)
  def find_key_path(self, key):
    查找key路径
    :param key: 需要查找路径的key值
    :return: 包含key值路径的list
    result = []
    for path,value in self.__paths(self.data):
      if path.endswith("['%s']" % key):
        result.append(path)
    with open('path.txt', 'w+', encoding='utf-8') as f:
      list(map(lambda line: f.write(line + '\r'), result))
    return result
  def find_value_path(self, key):
    查找某个值的路径
    :param key: 需要查找的值，限制为字符串，数字，浮点数，布尔值
    :return:
    result = []
    for path, value in self.__paths(self.data):
      if isinstance(value, (str, int, bool, float)):
        if value == key:
          result.append(path)
    with open('path.txt', 'w+', encoding='utf-8') as f:
      list(map(lambda line: f.write(line + '\r'), result))
    return result
if __name__ == '__main__':
  data = {'name': 'funcups'}
  hj = HandleJson(data)
  res = hj.find_key_path('name')
  print(res)
  res = hj.find_value_path('funcups')
  print(res)

补充拓展：python 获取的json字符串取值

获取到的json字符串，然后对其取值

{u’result’: {u’10.10.10.100′: {u’status’: u’OK’, u’msg’: u”{‘listen’: {”: set([]), ’25’: set([]), ’22’: set([‘10.9.19.148’, ‘10.10.10.1’, ‘10.10.10.130’])}}”}}}