Python Beautiful Soup .content Property
Solution 1:
It just gives you whats inside the tag. Let me demonstrate with an example:
html_doc = """
<html><head><title>The Dormouse's story</title></head>
<p class="title"><b>The Dormouse's story</b></p>
<p class="story">Once upon a time there were three little sisters; and their names were
<a href="http://example.com/elsie" class="sister" id="link1">Elsie</a>,
<a href="http://example.com/lacie" class="sister" id="link2">Lacie</a> and
<a href="http://example.com/tillie" class="sister" id="link3">Tillie</a>;
and they lived at the bottom of a well.</p>
<p class="story">...</p>
"""
from bs4 import BeautifulSoup
soup = BeautifulSoup(html_doc)
head = soup.head
print head.contents
The above code gives me a list,[<title>The Dormouse's story</title>], because thats inside the head tag. So calling [0] would give you the first item in the list.
The reason you get an error is because soup.contents[0].contents[0].contents[0].contents[0] returns something with no further tags (therefore no attributes). It returns Page Title from your code, because the first contents[0] gives you the HTML tag, the second one, gives you the head tag. The third one leads to the title tag, and the fourth one gives you the actual content. So, when you call a name on it, it has no tags to give you.
If you want the body printed, you can do the following:
soup = BeautifulSoup(''.join(doc))
print soup.body
If you want body using contents only, then use the following:
soup = BeautifulSoup(''.join(doc))
print soup.contents[0].contents[1].name
You will not get it using [0] as the index, because body is the second element after head.
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