beautifulsoup: get text (including html tags) between two different tags (</h3> and <h2>) - beautifulsoup

Question

I am trying to scrape an html file structured as follow using beautifulsoup. Basicaly, each unit is constisted of:
one <h2></h2>
one <h3></h3>
more than one <p></p>
Something like follow:
<h2>January, 2020</h2>
<h3>facility</h3>
<p>text1-1</p>
<p>text1-2</p>
<h2>April, 2020</h2>
<h3>scientists</h3>
<p>text2-1</p>
<p>text2-2</p>
<h2>June, 2020</h2>
<h3>lawyers</h3>
<p>text3-1</p>
<h2>.....
I want to get text including the <p> tags between </h3> and the next <h2>. The result should be:
for row #1:
<p>text1-1</p>
<p>text1-2</p>
for row #2:
<p>text2-1</p>
<p>text2-2</p>
for row #3:
<p>text3-1</p>
Here is what I tried so far:
num_h2 = len(soup.find_all('h2'))
for i in range(0,num_h2):
print('---------')
print(i)
p_string = ''
sibling = soup.find_all('h3')[i].find_next_sibling('p').getText()
if sibling:
p_string += sibling
else:
break
print(p_string)
The problem with this solution is that it only shows the content of the first <p> under each unit. I do not know how to find how many <p> are there to generate a for loop. Also, is there a better way to do this than using find_next_silibing()?

Answer 1

Maybe css selectors can help:
for s in soup.select('h3'):
for ns in (s.fetchNextSiblings()):
if ns.name == "h2":
break
else:
if ns.name == "p":
print(ns)
Output:
<p>text1-1</p>
<p>text1-2</p>
<p>text2-1</p>
<p>text2-2</p>
<p>text3-1</p>