How to cover all case of to with python and smart way?

Question

How to cover all case of <strong> <em> <u> to <strong> with python and smart way?

I was trying code to cover that but I think my way not good and have a lot of case.

Have any want know how to do that with better?

I very want have 2 way to learn that:

Way 1: Not use regex.
Way 2: Use regex.

My code now:

html = '''
<h1>Make bakery</h1>
<p>Step 1: Please use good quality product for make <strong><em>bakery</em></strong></p>
<p>Step 2: Make <em><u>bakery</u></em> hot</p>
<p>Step 2: Make <em><strong><u>bakery</u></strong></em> hot</p>
'''

def function_cover_all_strong(html):

    html = html.replace('<u><em><strong>','<strong>')
    html = html.replace('</strong></em></u>','</strong>')

    html = html.replace('<em><strong><u>','<strong>')
    html = html.replace('</u></strong></em>','</strong>')

    html = html.replace('<strong><u><em>','<strong>')
    html = html.replace('</em></u></strong>','</strong>')

    html = html.replace('<strong><em><u>','<strong>')
    html = html.replace('</u></em></strong>','</strong>')

    html = html.replace('<em><u>','<strong>')
    html = html.replace('</u></em>','</strong>')

    html = html.replace('<u><strong>','<strong>')
    html = html.replace('</strong></u>','</strong>')

    html = html.replace('<u><em>','<strong>')
    html = html.replace('</em></u>','</strong>')

    html = html.replace('<strong><u>','<strong>')
    html = html.replace('</u></strong>','</strong>')

    html = html.replace('<strong><em>','<strong>')
    html = html.replace('</em></strong>','</strong>')

    html = html.replace('<u>','<strong>')
    html = html.replace('</u>','</strong>')

    html = html.replace('<em>','<strong>')
    html = html.replace('</em>','</strong>')

    return html

html_cover = function_cover_all_strong(html = html)
print(html_cover)

Thanks for your support! All best with you. I very try to research more but not see case like that!

Obligatory https://stackoverflow.com/a/1732454/14637 – Thomas Nov 18 '21 at 15:41 — Thomas, Nov 18 '21 at 15:41
Thanks you so much @Thomas I will read that – nguyenphanhoaiduc Nov 19 '21 at 16:49 — nguyenphanhoaiduc, Nov 19 '21 at 16:49

alec_djinn · Accepted Answer · 2021-11-19T11:49:02.423

1

What about using a simple list and a for loop?

def function_cover_all_strong(html):

    #convert <strong class"some-thing" id="something"> to <strong>
    while '<strong class' in html:
        i = html.find('<strong class') #find start index
        j = html[i:].find('>') #find end index
        html = html[:i] + '<strong>' + html[j:] #replace


    #convert tags
    tags = ['<u><em><strong>', '</strong></em></u>', '<em><strong><u>']
    for old_tag in tags:
        if '/' in tag:
            new_tag = '</strong>'
        else:
            new_tag = '<strong>'

        html = html.replace(old_tag, new_tag)

    return html

edited Nov 19 '21 at 11:49

answered Nov 18 '21 at 15:38

alec_djinn

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Yes! That better for me! Thanks for you way. If it have id and class have any way to do that? Like this: Bakery. – nguyenphanhoaiduc Nov 18 '21 at 15:57
This case is very different from the one you posted. Can you post a few concrete examples? check https://pypi.org/project/beautifulsoup4/ for more complex HTML parsing. – alec_djinn Nov 18 '21 at 16:01
Yes! I need cover to . With some-thing is unknown keyword. Have any way to do that? – nguyenphanhoaiduc Nov 18 '21 at 16:04
I have updated the answer accordingly. – alec_djinn Nov 19 '21 at 11:44
Yes! Thanks you so much @alec_djinn I will try with your code. Thanks for your sharing! – nguyenphanhoaiduc Nov 19 '21 at 16:48

How to cover all case of to with python and smart way?

1 Answers1