Parse Output for Python

Question

My software outputs these two types of output:

-rwx------ Administrators/Domain Users  456220672   0% 2018-04-16 16:04:40 E:\\_WiE10-18.0.100-77.iso

-rwxrwx--- Administrators/unknown        6677   0% 2018-04-17 01:33:23 E:\\program files\\cluster groups\\sql server (mssqlserver)\\logs\\progress-MOD-1523883344023-3001-Windows.log

I would like to get the file names from both outputs:

E:\\_WiE10-18.0.100-77.iso, for the first one
E:\\program files\\cluster groups\\sql server (mssqlserver)\\logs\\progress-MOD-1523883344023-3001-Windows.log, for the second one

If i use something like the code below, it won't work if the second parameter has spaces in it. It works if there aren't any spaces in the Domain Username.

for item in outputs:
    outputs.extend(item.split())
for item2 in [' '.join(outputs[6:])]:
    new_list.append(item2)

How can I get all the parameters individually, including the filenames?

E. g. with a regular expression like \b[A-Z]:\\\\.+

Jan
– Jan

2018-04-22 12:23:57 +00:00
Commented Apr 22, 2018 at 12:23 — Jan
– Jan, Commented Apr 22, 2018 at 12:23

Patrick Artner · Accepted Answer · 2018-04-22 13:12:14Z

4

If regex is an option:

text = """-rwx------ Administrators/Domain Users  456220672   0% 2018-04-16 16:04:40 E:\\_WiE10-18.0.100-77.iso

-rwxrwx--- Administrators/unknown        6677   0% 2018-04-17 01:33:23 E:\\program files\\cluster groups\\sql server (mssqlserver)\\logs\\progress-MOD-1523883344023-3001-Windows.log"""

import re

for h in re.findall(r"^.*?\d\d:\d\d:\d\d (.*)",text,flags=re.MULTILINE):
    print(h)

Output:

E:\_WiE10-18.0.100-77.iso
E:\program files\cluster groups\sql server (mssqlserver)\logs\progress-MOD-1523883344023-3001-Windows.log

Pattern explained:

The pattern r"^.*?\d\d:\d\d:\d\d (.*)" looks for linestart '^' + as less anythings as possible '.*?' + the time-stamp '\d\d:\d\d:\d\d ' followed by a space and captures all behind it till end of line into a group.

It uses the re.MULTILINE flag for that.

Edit:

Capturing the individual things needs some more capturing groups:

import re

for h in re.findall(r"^([rwexXst-]+) ([^0-9]+) +\d+.+? +(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}) (.*)",text,flags=re.MULTILINE):
#                       ^^^^^^^^^^^^ ^^^^^^^^           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^
#                          flags     grpName                               datetime          filename
    for k in h:
        print(k)
    print("")

Output:

-rwx------
Administrators/Domain Users 
2018-04-16 16:04:40
E:\_WiE10-18.0.100-77.iso

-rwxrwx---
Administrators/unknown       
2018-04-17 01:33:23
E:\program files\cluster groups\sql server (mssqlserver)\logs\progress-MOD-1523883344023-3001-Windows.log

edited Apr 22, 2018 at 13:12

answered Apr 22, 2018 at 12:31

Patrick Artner

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8 Comments

PanDe Over a year ago

How can i get all parameters individually like if i only need permissions(first one) and date and time as well?

PanDe Over a year ago

i got the first one : for fileperm in re.findall(r"([^ ]+) .*",backup_files_val,flags=re.MULTILINE): print(fileperm)

Patrick Artner Over a year ago

@PetPan you would need more capturing groups. Look: regexr.com/3oart .. something along '^([rwexXst-]+) ([^0-9]+) +\d+.+? +(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}) (.*)' - groups 1 to 4 will give you the different capturing groups.

Patrick Artner Over a year ago

@PetPan look at the regex - the filesize is currently not captured: change regex to re.findall(r"^([rwexXst-]+) ([^0-9]+) +(\d+) .+? +(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}) (.*)",text,flags=re.MULTILINE):

PanDe Over a year ago

i Slightly modified it to get the full filesize, your regex was missing + sign for the following ^([rwexXst-]+) ([^0-9]+) +(\d+).... now, its working, thanks again. cheers

|

Jan · Accepted Answer · 2018-04-22 12:27:19Z

2

You could use a regular expression like

\b[A-Z]:\\\\.+

answered Apr 22, 2018 at 12:27

Jan

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Comments

Sean Francis N. Ballais · Accepted Answer · 2018-04-23 01:49:44Z

1

Aside from using regex, you can try something similar to this.

output = '-rwx------ ... 2018-04-16 16:04:40 E:\\\\_WiE10-18.0.100-77.iso'

drive_letter_start = output.find(':\\\\')
filename = output[drive_letter_start - 1:]

It looks for the first occurrence of ':\\'and gets the drive letter before the substring (i.e. ':\\') and the full file path after the substring.

EDIT
Patrick Artner's answer is better and completely answers OP's question compared to this answer. This only encompasses capturing the file path. I am leaving this answer here should anyone find it useful.

edited Apr 23, 2018 at 1:49

answered Apr 22, 2018 at 12:25

Sean Francis N. Ballais

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