Python: filename matching using regex

Question

I have written this code

import os
from datetime import datetime
import re

now = datetime.now() 
filename = now.strftime("%Y%m%d%H%M") #For example 202006191839

for fname in os.listdir(downloadPath):                
   if re.match('export_' + filename + '[0-9]{2}.xlsx', fname):
      print(fname)

In downloadPath I have these files

export_20200619183900.xlsx
export_20200619183921.xlsx
export_20200619183930.xlsx

But the re.match is not matching as desired.

But, if i change

filename = now.strftime("%Y%m%d%H%M")

with a simple assignment

filename = "202006191839"

The code works. The problem is, I need to have dynamic data.

Can anyone help me?

Errrr, `datetime.now()` gives your current, local time. Are you sure you have a file with this timestamp? It will change every time you run your script. — bavaza, Jun 19 '20 at 16:53
Yes, the file is download in the same procedure, a few seconds earlier. — SirLancillotto, Jun 19 '20 at 17:08

score 1 · Answer 1 · answered Jun 19 '20 at 16:58

1

I think it is because you are matching 'export_' + filename, but you said the file was excel_20200619183900

answered Jun 19 '20 at 16:58

Vthechamp

636
1
8
20

No no, I am sorry, the file is export_xxxxxxxx. I was wrong to write in the post! – SirLancillotto Jun 19 '20 at 17:10

score 0 · Accepted Answer · answered Jul 03 '20 at 07:30

0

Ok.

I am solved the problem... i am blind probably The file I search, is download before the above code, despite being very small, the search command starts before the download... I have add a simple time.sleep(2) before search command.

Thanks to all.

answered Jul 03 '20 at 07:30

SirLancillotto

49
8

Python: filename matching using regex

2 Answers2