Python regex fetch the date form log file

Question

I want to fetch the date from the standard vsftp log file, the file is as below:

Fri Sep  1 15:43:46 2017 1 ::ffff:172.18.1.168 14848 /IT_inventory.xls b _ i r user01 ftp 0 * c

I tried to use regex:

#!/usr/bin/python
import re
    with open("/var/log/xferlog") as log:
        for line in log:
            pattern = re.compile(r'(\w+) (\w+)')
            match = pattern.search(line)
            print match.group(1)
            print match.group(2)

The code can fetch day and month ( group(1) = Fri, group(2) = Sep ) correctly.

But I want to fetch the date(1), so I change the pattern:

pattern = re.compile(r'(\w+) (\w+) (\d+)')
        match = pattern.search(line)
        print match.group(1)
        print match.group(2)
        print match.group(3)

But the output change to 46, 2017, 1 ( group(1) = 46, group(2) = 2017, group(3) = 1 )

If I set up line manually.

line = "Fri Sep 1 15:43:46 2017 1 ::ffff:172.18.1.168 14848 /IT_inventory.xls b _ i r user01 ftp 0 * c"

pattern = re.compile(r'(\w+) (\w+) (\d+)')
match = pattern.search(line)
print match.group(1)
print match.group(2)
print match.group(3)

Then I can get what I want.

group(1)=Fri, group(2)=Sep, group(3)=1

Can anyone explain it to me?

Ajax1234 · Accepted Answer · 2017-09-01 23:29:07Z

2

You can try this:

import re

s = "Fri Sep  1 15:43:46 2017 1 ::ffff:172.18.1.168 14848 /IT_inventory.xls b _ i r user01 ftp 0 * c"

data = re.findall("^\w+\s\w+", s)[0].split()

Output:

['Fri', 'Sep']

answered Sep 1, 2017 at 23:29

Ajax1234

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1 Comment

ITnewbie Over a year ago

Thanks for your solution! It worked. But I am still very curious. Do you know why I will get the wired result if I read the file by syntax "with"?

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