Python beautiful soup syntax error

Question

I'm using an example I found that extracts links from an webpage

from BeautifulSoup import BeautifulSoup
import urllib2
import re

html_page = urllib2.urlopen("http://arstechnica.com")
soup = BeautifulSoup(html_page)
for link in soup.findAll('a', attrs={'href': re.compile("^http://")}):
print link.get('href')

I get:

File "html_test.py", line 9
    print link.get('href')
             ^
SyntaxError: invalid syntax

I've installed BeautifulSoup with pip to no success.

What python version do you use. If python 3, then `print` should be a function: `print(...)`. — Yevhen Kuzmovych, Jan 19 '18 at 01:23
In python 3, `print` is not a keyword, rather than a function. So you should do `print(link.get('href'))`. — kyriakosSt, Jan 19 '18 at 01:24
And `urllib2` is `urllib.request`. And it's best to use `bs4` instead of `BeautifulSoup` — t.m.adam, Jan 19 '18 at 01:25
Possible duplicate of [Syntax error on print with Python 3](https://stackoverflow.com/questions/826948/syntax-error-on-print-with-python-3) — OneCricketeer, Jan 19 '18 at 01:29
I may need to update Ubuntu 17.04 to 17.10, because I can't seem to even update packages.... — , Jan 19 '18 at 01:31

score 1 · Answer 1 · edited Jan 19 '18 at 01:30

1

Instead of

from BeautifulSoup import BeautifulSoup

use:

from bs4 import BeautifulSoup

Also, in Python3, print is a function, not a statement:

print(link.get('href'))

edited Jan 19 '18 at 01:30

OneCricketeer

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answered Jan 19 '18 at 01:24

Ajax1234

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Python beautiful soup syntax error

1 Answers1