CODEWITHSUNDEEP
pythonbeautifulsoup
Jason Christa
Jason Christa

Reputation: 12508

Remove a tag using BeautifulSoup but keep its contents

Currently I have code that does something like this:

soup = BeautifulSoup(value)

for tag in soup.findAll(True):
    if tag.name not in VALID_TAGS:
        tag.extract()
soup.renderContents()

Except I don't want to throw away the contents inside the invalid tag. How do I get rid of the tag but keep the contents inside when calling soup.renderContents()?

Upvotes: 64

Views: 69354

Answers (12)

Thom Ives
Thom Ives

Reputation: 3989

What Worked For Me On Python 3.10 With BS4 And Unwrap

I initially liked Jesse Dhillon's answer a lot. However, I kept running into issues with the recursive calls due to recalling of the parser in BS4. I tried to change the level of recursion, but I kept running into problems with that too.

Then I looked into applying Bishwas Mishra's answer. Due to changes in BS4, I had to modify his code a bit, and I finally was able to develop a piece of code that would remove tags and maintain content.

I hope this helps some others.

from bs4 import BeautifulSoup


html = "<p>Good, <b>bad</b>, and <i>ug<b>l</b><u>y</u></i></p>"

soup = BeautifulSoup(html, "html5lib")

for c in ["html", "head", "body", "b", "i", "u"]:
    while soup.find(c):
        exec(f"soup.{c}.unwrap()")

print(soup)

NOTE: It is necessary to add "html", "head", and "body" to the invalid tags list, because BS4 will add those into your html text if they were not originally there, and I did not want them for my specific case.

The output I got from the above code was ...

<p>Good, bad, and ugly</p>

Upvotes: 1

Dom DaFonte
Dom DaFonte

Reputation: 1779

Here is a python 3 friendly version of this function: 

from bs4 import BeautifulSoup, NavigableString
invalidTags = ['br','b','font']
def stripTags(html, invalid_tags):
    soup = BeautifulSoup(html, "lxml")
    for tag in soup.findAll(True):
        if tag.name in invalid_tags:
            s = ""
            for c in tag.contents:
                if not isinstance(c, NavigableString):
                    c = stripTags(str(c), invalid_tags)
                s += str(c)
            tag.replaceWith(s)
    return soup

Upvotes: 2

Bishwas Mishra
Bishwas Mishra

Reputation: 1342

Use unwrap.

Unwrap will remove one of multiple occurrence of the tag and still keep the contents.

Example:

>> soup = BeautifulSoup('Hi. This is a <nobr> nobr </nobr>')
>> soup
<html><body><p>Hi. This is a <nobr> nobr </nobr></p></body></html>
>> soup.nobr.unwrap
<nobr></nobr>
>> soup
>> <html><body><p>Hi. This is a nobr </p></body></html>

Upvotes: 3

robus gauli
robus gauli

Reputation: 391

Here is the better solution without any hassles and boilerplate code to filter out the tags keeping the content.Lets say you want to remove any children tags within the parent tag and just want to keep the contents/text then,you can simply do:

for p_tags in div_tags.find_all("p"):
    print(p_tags.get_text())

That's it and you can be free with all the br or i b tags within the parent tags and get the clean text.

Upvotes: 2

Tommz
Tommz

Reputation: 3453

This is an old question, but just to say of a better ways to do it. First of all, BeautifulSoup 3* is no longer being developed, so you should rather use BeautifulSoup 4*, so called bs4.

Also, lxml has just function that you need: Cleaner class has attribute remove_tags, which you can set to tags that will be removed while their content getting pulled up into the parent tag.

Upvotes: 1

corford
corford

Reputation: 1035

Although this has already been mentoned by other people in the comments, I thought I'd post a full answer showing how to do it with Mozilla's Bleach. Personally, I think this is a lot nicer than using BeautifulSoup for this.

import bleach
html = "<b>Bad</b> <strong>Ugly</strong> <script>Evil()</script>"
clean = bleach.clean(html, tags=[], strip=True)
print clean # Should print: "Bad Ugly Evil()"

Upvotes: 21

jimmy
jimmy

Reputation: 112

you can use soup.text

.text removes all tags and concatenate all text.

Upvotes: 7

Olof Sj&#246;bergh
Olof Sj&#246;bergh

Reputation: 73

None of the proposed answered seemed to work with BeautifulSoup for me. Here's a version that works with BeautifulSoup 3.2.1, and also inserts a space when joining content from different tags instead of concatenating words. 

def strip_tags(html, whitelist=[]):
    """
    Strip all HTML tags except for a list of whitelisted tags.
    """
    soup = BeautifulSoup(html)

    for tag in soup.findAll(True):
        if tag.name not in whitelist:
            tag.append(' ')
            tag.replaceWithChildren()

    result = unicode(soup)

    # Clean up any repeated spaces and spaces like this: '<a>test </a> '
    result = re.sub(' +', ' ', result)
    result = re.sub(r' (<[^>]*> )', r'\1', result)
    return result.strip()

Example:

strip_tags('<h2><a><span>test</span></a> testing</h2><p>again</p>', ['a'])
# result: u'<a>test</a> testing again'

Upvotes: 2

Jesse Dhillon
Jesse Dhillon

Reputation: 7997

The strategy I used is to replace a tag with its contents if they are of type NavigableString and if they aren't, then recurse into them and replace their contents with NavigableString, etc. Try this:

from BeautifulSoup import BeautifulSoup, NavigableString

def strip_tags(html, invalid_tags):
    soup = BeautifulSoup(html)

    for tag in soup.findAll(True):
        if tag.name in invalid_tags:
            s = ""

            for c in tag.contents:
                if not isinstance(c, NavigableString):
                    c = strip_tags(unicode(c), invalid_tags)
                s += unicode(c)

            tag.replaceWith(s)

    return soup

html = "<p>Good, <b>bad</b>, and <i>ug<b>l</b><u>y</u></i></p>"
invalid_tags = ['b', 'i', 'u']
print strip_tags(html, invalid_tags)

The result is:

<p>Good, bad, and ugly</p>

I gave this same answer on another question. It seems to come up a lot.

Upvotes: 64

Etienne
Etienne

Reputation: 12590

I have a simpler solution but I don't know if there's a drawback to it.

UPDATE: there's a drawback, see Jesse Dhillon's comment. Also, another solution will be to use Mozilla's Bleach instead of BeautifulSoup.

from BeautifulSoup import BeautifulSoup

VALID_TAGS = ['div', 'p']

value = '<div><p>Hello <b>there</b> my friend!</p></div>'

soup = BeautifulSoup(value)

for tag in soup.findAll(True):
    if tag.name not in VALID_TAGS:
        tag.replaceWith(tag.renderContents())

print soup.renderContents()

This will also print <div><p>Hello there my friend!</p></div> as desired.

Upvotes: 11

slacy
slacy

Reputation: 11783

Current versions of the BeautifulSoup library have an undocumented method on Tag objects called replaceWithChildren(). So, you could do something like this:

html = "<p>Good, <b>bad</b>, and <i>ug<b>l</b><u>y</u></i></p>"
invalid_tags = ['b', 'i', 'u']
soup = BeautifulSoup(html)
for tag in invalid_tags: 
    for match in soup.findAll(tag):
        match.replaceWithChildren()
print soup

Looks like it behaves like you want it to and is fairly straightforward code (although it does make a few passes through the DOM, but this could easily be optimized.)

Upvotes: 86

Alex Martelli
Alex Martelli

Reputation: 882681

You'll presumably have to move tag's children to be children of tag's parent before you remove the tag -- is that what you mean?

If so, then, while inserting the contents in the right place is tricky, something like this should work:

from BeautifulSoup import BeautifulSoup

VALID_TAGS = 'div', 'p'

value = '<div><p>Hello <b>there</b> my friend!</p></div>'

soup = BeautifulSoup(value)

for tag in soup.findAll(True):
    if tag.name not in VALID_TAGS:
        for i, x in enumerate(tag.parent.contents):
          if x == tag: break
        else:
          print "Can't find", tag, "in", tag.parent
          continue
        for r in reversed(tag.contents):
          tag.parent.insert(i, r)
        tag.extract()
print soup.renderContents()

with the example value, this prints <div><p>Hello there my friend!</p></div> as desired.

Upvotes: 7

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