Get text of next sibling based on text of previous sibling

Question

I have the following HTML:

<div id="infoTable">
    <h4>
      User
    </h4>
    <table>
        <tbody>
            <tr>
                <td class="name">
                    <a href="/userpage/123">BillyBob12345</a>
                </td>
            </tr>
            <tr>
                <td class="name">
                    <a href="/userpage/124">JimBob43</a>
                </td>
            </tr>
        </tbody>
    </table>
    <h4>
      Super User
    </h4>
    <table>
        <tbody>
            <tr>
                <td class="name">
                    <a href="/userpage/112">CookieMonster</a>
                </td>
            </tr>
        </tbody>
    </table>
</div>

Basically, I am looking to get two lists:

Users = [{"BillyBob12345" : "123"}, {"JimBob43" : "124"}]
SuperUsers = [{"CookieMonster" : "112"}]

I am currently using Python 2.7 with BeautifulSoup4 and I am able to find all of the users, but I can't split them up into their respectful groups.

Answer 1

If you happen to know that they are in that order, you could just use a list comprehension to create those lists of dictionaries, parsing the "userpage" number using .split('/'):

firstTable = soup.findAll('table')[0]
users = [{a.text : a['href'].split('/')[2]} for a in firstTable.findAll('a')]

secondTable = soup.findAll('table')[1]
superUsers = [{a.text : a['href'].split('/')[2]} for a in secondTable.findAll('a')]

---

>>> users
[{'BillyBob12345': '123'}, {'JimBob43': '124'}]
>>> superUsers
[{'CookieMonster': '112'}]

---

If you want to access the name "Users" to then use it into a dictionary, you can use:

>>> firstTable.previousSibling.previousSibling
<h4>
      User
    </h4>

Answer 2

I was actually able to extract the info using this:

if (BS.find('div').find('h4',text="User")):
    FindUsers = BS.find('div').find('h4', text="User").findNext('table').find('td', {"class" : "name"}).findAll('a')
    Users = [{u.text.strip() :  u['href'].split('/')[2]} for u in FindUsers ]