遍历”Day1-homework”目录下文件；

找到文件名包含“2020”的文件；

将文件名保存到数组result中；

按照序号、文件名分行打印输出。

 1 import os
 2 #待搜索的目录路径
 3 path = "Day1-homework"
 4 #待搜索的名称
 5 filename = "2020"
 6 #定义保存结果的数组
 7 result = []
 8 
 9 def findfiles(files_path, files_list):
10     #查找文件代码
11     files = os.listdir(files_path)
12     for s in files:
13         s_path = os.path.join(files_path, s)
14         if os.path.isdir(s_path):
15             findfiles(s_path, files_list)
16         elif os.path.isfile(s_path) and '2020' in s:
17             result.append(s_path)
18 
19 
20 if __name__ == '__main__':
21     findfiles(path,result)
22     for i in range(len(result)):
23         print("[{} ,".format(i)+"'"+result[i]+"\']")

输出结果：

[0 ,'Day1-homework/26/26/new2020.txt']
[1 ,'Day1-homework/18/182020.doc']
[2 ,'Day1-homework/4/22/04:22:2020.txt']