如何使用python从内联样式标签中删除特定的值对？

Question

i am trying to parse some html that has some nasty inline styling. it looks something like this

<span class="text_line" data-complex="0" data-endposition="4:2:86:5:0" data-position="4:2:74:2:0" style="font-family: scala-sans-offc-pro--; width: 100%; word-spacing: -2.66667px; font-size: 24px !important; line-height: 40px; font-variant-ligatures: common-ligatures; display: block; height: 40px; margin-left: 75px; margin-right: 155px;">

I am trying to remove just the attribute-value pair word-spacing: -2.66667px; . Here is the catch there are several hundred of these lines and no two are the same. Sometimes the spacing is word-spacing: -4px and sometimes word-spacing: -3.78632px; or some other random number.

I tried the beautiful soup, I figured out how to remove the whole tag, which is not what I wanted. I don't know how to do it with regular expressions. And I read that it's better to avoid trying to edit HTML with regex.

My idea right constitutes saving all the span tags to a variable using beautiful soup and then using string.find() to get the indexes of all the "w"'s in word-spacing and then finding the next semi column. Then after I have a list find a way to cut the string at those indexes and join the remnants back together. Maybe splitting at the ";" is better... I don't know any more at this point. The brain is a fried and tired. :P

    def __init__(self, first_index, last_index):
        self.first = first_index
        self.last = last_index
def getIndices(text, start_index):
    index = CutPointIndex(None, None)
    index.first = text.find("word-spacing", start_index, end_index)
    if(index.first != -1):
        index.last = text.find(";", index.first , end_index)
    return index

Given something like style="font-family: scala-sans-offc-pro--; width: 100%; word-spacing: -3.71429px;"

or style="font-family: scala-sans-offc-pro--; width: 100%; word-spacing: -5px;

or any other variation of values the expected outcome should be style="font-family: scala-sans-offc-pro--; width: 100%;

Answer 1

I'm guessing that maybe you might want to re.sub the variable word-spacing :

import re

regex = r"\s*word-spacing\s*:\s*[^;]*;"

test_str = '''
style="font-family: scala-sans-offc-pro--; width: 100%; word-spacing: -3.71429px;"
style="font-family: scala-sans-offc-pro--; width: 100%; word-spacing: -5px;"
style="font-family: scala-sans-offc-pro--; width: 100%;"

'''

print(re.sub(regex, "", test_str))

---

Output

style="font-family: scala-sans-offc-pro--; width: 100%;"
style="font-family: scala-sans-offc-pro--; width: 100%;"
style="font-family: scala-sans-offc-pro--; width: 100%;"

---

If you wish to explore/simplify/modify the expression, it's been explained on the top right panel of regex101.com . If you'd like, you can also watch in this link , how it would match against some sample inputs.

---

Answer 2

You could match on elements with that attribute and remove that part.

I split the style attribute (for the relevant tags only) on ; then recombine excluding the unwanted pair

';'.join([i for i in t['style'].split(';') if 'word-spacing' not in i])

but you could just as easily update the value for word-spacing

from bs4 import BeautifulSoup as bs

html = '''
<span class="text_line" data-complex="0" data-endposition="4:2:86:5:0" data-position="4:2:74:2:0" style="font-family: scala-sans-offc-pro--; width: 100%; word-spacing: -2.66667px; font-size: 24px !important; line-height: 40px; font-variant-ligatures: common-ligatures; display: block; height: 40px; margin-left: 75px; margin-right: 155px;">
'''
soup = bs(html, 'lxml')

for t in soup.select('[style*= word-spacing]'):
    t['style'] = ';'.join([i for i in t['style'].split(';') if 'word-spacing' not in i])
print(soup)

Reading:

1. https://www.crummy.com/software/BeautifulSoup/bs4/doc/#attributes
2. https://developer.mozilla.org/en-US/docs/Web/CSS/Attribute_selectors