如何逐行打印在线txt文件？

Question

My codes are listed as following:

import urllib.request, urllib.parse, urllib.error

headers = {'User-Agent':
                       '''Mozilla/5.0 (Windows NT 10.0; Win64; x64) 
                       AppleWebKit/537.36 (KHTML, like Gecko) 
                       Chrome/79.0.3945.117 Safari/537.36'''}

url = 'https://www.py4e.com/code3/mbox.txt'

req = urllib.request.Request(url, headers=headers)

fname = urllib.request.urlopen(req).read().decode()

for line in fname:
    print(line)

The output is not line by line, but characters? How can I print the file line by line? Thanks a lot!

Answer 1

Try using split by newline for your string.

for line in fname.split('\n'):
    print(line)

Answer 2

Your read() is reading the text into a string.

import urllib.request, urllib.parse, urllib.error

headers = {'User-Agent':
                       '''Mozilla/5.0 (Windows NT 10.0; Win64; x64) 
                       AppleWebKit/537.36 (KHTML, like Gecko) 
                       Chrome/79.0.3945.117 Safari/537.36'''}

url = 'https://www.py4e.com/code3/mbox.txt'

req = urllib.request.Request(url, headers=headers)

fname = urllib.request.urlopen(req)
count = 0
for line in fname:
    count += 1
    print(count)
    print(line.decode())

Which Yields:

1
From stephen.marquard@uct.ac.za Sat Jan  5 09:14:16 2008

2
Return-Path: <postmaster@collab.sakaiproject.org>

3
Received: from murder (mail.umich.edu [141.211.14.90])

4
by frankenstein.mail.umich.edu (Cyrus v2.3.8) with LMTPA;

5     
Sat, 05 Jan 2008 09:14:16 -0500

6
X-Sieve: CMU Sieve 2.3

7
Received: from murder ([unix socket])

8
by mail.umich.edu (Cyrus v2.2.12) with LMTPA;

9
Sat, 05 Jan 2008 09:14:16 -0500

10
Received: from holes.mr.itd.umich.edu (holes.mr.itd.umich.edu [141.211.14.79])
...