Parsing données de la table HTML en utilisant le dictionnaire python

Question

Je reçois le code html de la page Jive ci-dessous le format

table = <table class="test" style="border: 1px solid #c6c6c6;" width="100%"><thead><tr><th style="background-color: #efefef; width: 13%;">Tag</th><th style="background-color: #efefef; width: 23.7965%;">ID</th><th style="background-color: #efefef; width: 59.2035%;">URL</th></tr></thead><tbody><tr><td style="width: 13%;">3.7.3</td><td style="width: 23.7965%;"><p>12345</p><p>232323</p><p>4343454</p><p>5454554</p></td><td style="width: 59.2035%;"><p><a class="jive-link-external-small" href="http://google.com" rel="nofollow">http://google.com</a></p><p><a class="jive-link-external-small" href="http://test123.com" rel="nofollow">http://test123.com</a></p><p><a class="jive-link-external-small" href="http://www.yahoo.com" rel="nofollow">http://www.yahoo.com</a></p><p><a class="jive-link-external-small" href="http://www.test.com" rel="nofollow">http://www.test.com</a></p></td></tr><tr><td style="width: 13%;">3.7.4</td><td style="width: 23.7965%;"><p>456789</p><p>545454</p><p>5454545</p><p>545454</p></td><td style="width: 59.2035%;"><p><a class="jive-link-external-small" href="http://foo.com" rel="nofollow">http://foo.com</a></p><p><a class="jive-link-external-small" href="http://www.yahoo.com" rel="nofollow">http://www.yahoo.com</a></p><p><a class="jive-link-external-small" href="http://svn.com" rel="nofollow">http://svn.com</a></p><p><a class="jive-link-external-small" href="http://test.com" rel="nofollow">http://test.com</a></p></td></tr></tbody></table> 

de la devise HTML à un dictionnaire j'ai essayé ci-dessous le code

table = ET.XML(s) 
rows = iter(table) 
headers = [col.text for col in next(rows)] 
for row in rows: 
    values = [col.text for col in row] 
    out = dict(zip(headers, values)) 

L'approche ci-dessus ne donnez-moi la sortie attendue comme ci-dessous basé sur l'entrée de l'argument de ligne de commande

Tag  ID   URL 
3.7.3  121211  http://yahoo.com 
      323243  http://url.com 

Answer 1

Ceci peut être réalisé usin g Arborescence XML Python.

data = response.json() 
    html_doc = data['content']['text'] 

def find_version(ver): 
    table = ET.XML(html_doc) 

    # headers 
    ths = [th.text for th in table.findall('.//th')] 

    for tr in table.findall('.//tbody/tr'): 
     data = [] 

     # first col 
     data.append(tr[0].text) 

     # second col 
     data.append([x.text for x in tr[1]]) 

     # third col 
     temp = [] 
     for x in tr[2]: 
      if x.tag == 'a': 
       temp.append(x.text) 
      else: 
       temp.append(x[0].text) 
     data.append(temp) 

     # dictionary to print all the available data using applicable keys 
     #out = OrderedDict(zip(ths, data)) 
     out = OrderedDict(zip(ths, data)) 
     #print('out:', out) 

     if out['Release'] == ver: 
      return out 

# --- main --- 

res = find_version(ver) 

if res: 
    ''' 
    for key, val in res.items(): 
     print(key, '-', val) 
    ''' 
    #print (res.values()) 
    f = lambda x: x if isinstance(x, list) else [x] 

    for tup in zip(*res.items()): 
     for a, b, c in zip_longest(*map(f, tup), fillvalue=''): 
      print('{:15}{:15}{:15}'.format(a, b, c)) 
     #for a, b, c in zip_longest(*map(f, tup), fillvalue=''): 
      #print('{:15}'.format(b))     
else: 
    print ('Version not found') 

vous donnera la sortie au format prévu