How do I avoid the 'NavigableString' error with BeautifulSoup and get to the text of href? - beautifulsoup

Question

This is what I have:
import requests
from bs4 import BeautifulSoup
from urllib.parse import urljoin
url = "http://python.beispiel.programmierenlernen.io/index.php"
doc = requests.get(url).content
soup = BeautifulSoup(doc, "html.parser")
for i in soup.find("div", {"class":"navigation"}):
print(i)
Currently the print output of "i" is:
<a class="btn btn-primary" href="index.php?page=2">Zur nächsten Seite!</a>
I want to print out the href link "index.php?page=2".
When I try to use BeautifulSoups "find", "select" or "attrs" method on "i" I get an error. For instance with
print(i.attrs["href"])
I get:
AttributeError: 'NavigableString' object has no attribute 'attrs'
How do I avoid the 'NavigableString' error with BeautifulSoup and get the text of href?

Answer 1

The issue seems to be for i in soup.find. If you're looking for only one element, there's no need to iterate that element, and if you're looking for multiple elements, find_all instead of find would probably match the intent.
More concretely, here are the two approaches. Beyond what's been mentioned above, note that i is a div that contains the desired a as a child, so we need an extra step to reach it (this could be more direct with an xpath).
import requests
from bs4 import BeautifulSoup
url = "http://python.beispiel.programmierenlernen.io/index.php"
doc = requests.get(url).content
soup = BeautifulSoup(doc, "html.parser")
for i in soup.find_all("div", {"class": "navigation"}):
print(i.find("a", href=True)["href"])
print(soup.find("div", {"class": "navigation"})
.find("a", href=True)["href"])
Output:
index.php?page=2
index.php?page=2