Convert streamlit markdown output to a string without losing HTML and CSS styling

Using Streamlit
1

My Streamlit app is deployed locally (i.e., testing phase). However, I have created an app that first and foremost, collect data from a web page using BeautifulSoup “web scraping method”. Then this data is displayed as a dataframe in a table format that allows the user to select a list of items in the displayed table as shown below :

import requests, smtplib
import pandas as pd, streamlit as st, datetime
from bs4 import BeautifulSoup as bs
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def main():
st.set_page_config(
        page_title="Web Scraping Data",
        layout="wide",
        initial_sidebar_state="auto")
    st.header("Data")
    cur_date = datetime.datetime.now()
    formatted_date = cur_date.strftime('%d/%m/%Y')
    st.write("Date", cur_date)
    # get the content of URL using request module and store it in page variable
    # create a soup object from HTML data
    URL = "https://liphy.univ-grenoble-alpes.fr/fr/actualites"
    page = requests.get(URL).content
    soup = bs(page,'html.parser')

    actualite = soup.find("ul", {"class":"liste__objets liste__actualites flex-oui flex-2-3"})
    # extract all titles, contents, date and links
    pub_date = []
    titre = []
    contente = []
    lien = []
    # link is not used here but can be extracted if needed later on for further analysis purposes
    for lst in actualite.find_all("li", {"class":"liste__objets__style0004"}):
        title = lst.find("em").get_text(strip=True) # extracting title
        content = lst.find("div", {"class":"liste__objets__resume"}, recursive=True).get_text(strip=True)
        # Iterate through each link and extract its href attribute and text
        link = lst.find("a", href=True)["href"] 
        pub_date.append(date)
        titre.append(title)
        contente.append(content)
        lien.append(link)
    df = pd.DataFrame({'Date':pub_date, 'Titre':titre, 'Résumé':contente, 'Lien':lien})
    df1 = df.replace(r'\r+|\n+|\t+','', regex=True)
    selected_rows = selected_rows_from_dataframe(df1)
    st.divider() 
    st.subheader("Transform")
    container = st.container() 


for index, row in selected_rows.iterrows():
         with container:
            #   write the message header and semi-text
              st.markdown(f'**<p style="font-size:20px;">{row["Titre"]}</p>**', unsafe_allow_html=True, help=None)
              st.markdown(f'{row["Résumé"]}')
              st.markdown(f'<a href="{row["Lien"]}" target="_blank" style="font-size:15px;">Lire la suite</a>', unsafe_allow_html=True)


if __name__ == "__main__":
    main()

From the code above, the second section that begins with “for index” i would like to convert the st.markdown’s output as a string without losing the formatted text and send it as a mail.
Thank you for your input and in case there is a better alternative to achieve this, I would welcome all suggestions

2

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