Python遍历文件夹 处理json文件的方法
有两种做法:os.walk()、pathlib库,个人感觉pathlib库的path.glob用来匹配文件比较简单。
下面是第二种做法的实例(第一种做法百度有很多文章):
from pathlib import Path import json analysis_root_dir = "D:\\analysis_data\json_file" store_result="D:\\analysis_data\\analysis_result\\dependency.csv" def parse_dir(root_dir): path = Path(root_dir) all_json_file = list(path.glob('**/*.json')) parse_result = [] for json_file in all_json_file: # 获取所在目录的名称 service_name = json_file.parent.stem with json_file.open() as f: json_result = json.load(f) json_result["service_name"] = service_name parse_result.append(json_result) return parse_result def write_result_in_file(write_path , write_content): with open(write_path,'w') as f: f.writelines("service_name,action,method,url\n") for dict_content in write_content: url = dict_content['url'] method = dict_content['method'] action = dict_content['action'] service_name = dict_content['service_name'] f.writelines(service_name + ","+ action+","+method + ","+ url+"\n") def main(): print("main begin...") parse_result = parse_dir(analysis_root_dir) print(parse_result) write_result_in_file(store_result,parse_result) print("main finished...") if __name__ == '__main__': main()
运行结果
main begin... [{'url': '/rest/webservice/v1/dosomthing', 'method': 'post', 'action': 'create', 'service_name': 'WebSubService'}, {'url': '/rest/webservice/v1/dosomthing', 'method': 'post', 'action': 'create', 'service_name': 'WebSubService01'}, {'url': '/rest/webservice/v1/dosomthing', 'method': 'post', 'action': 'create', 'service_name': 'WebSubService02'}, {'url': '/rest/webservice/v1/dosomthing', 'method': 'post', 'action': 'create', 'service_name': 'WebSubService03'}, {'url': '/rest/webservice/v1/dosomthing', 'method': 'post', 'action': 'create', 'service_name': 'WebSubService04'}, {'url': '/rest/webservice/v1/dosomthing', 'method': 'post', 'action': 'create', 'service_name': 'WebSubService05'}] main finished...
目录结构
json file内容
{ "url":"/rest/webservice/v1/dosomthing", "method":"post", "action":"create" }
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