print start of html tags

Question

I want to print out the first html tags thats has attributes

    
test
    
test2
    


        
                      

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Answer 1

This seems pretty complicated, you can try with this expression, but it would fail in some cases. It would first collect the undesired instances, then at the end there is a capturing group for those desired.

Maybe, it wouldn't be the best idea to use regular expressions here.

Test

import re

regex = r"^\s*<\S+>\s*$|^\s*<\S+\s.*test.*?>.*?<\/\S+>$|^\s*(<.*>)\s*$"

test_str = """

<h1>test</h1>
    <h2>test2</h2>
    <div id="content"></div>
    <p>test3</p>
    <div class="test"></div>
    <div id="nav"></div>
    <p>test3</p>

"""

print(re.findall(regex, test_str, re.M))

Output

['', '', '<div id="content"></div>', '', '', '<div id="nav"></div>', '']

The expression is explained on the top right panel of regex101.com, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs, if you like.

Answer 2

You should use a non-greedy match for any number of characters to the left of the =, so:

r'<.*?=.*?>'

That will match a <, followed by a minimum number of characters, followed by a =, followed by the minimum number of characters until the >.

What you had:

r'<?=.*?>'

Means an optional <, followed by a =, followed by any string going up to the >. Since the < is optional and would only match if right before the =, you end up with no matches for it.