使用BeautifulSoup获取第n个元素

Answer 1

您还可以使用findAll获取列表中的所有行，然后使用slice语法访问所需的元素：

rows = soup.findAll('tr')[4::5]

Answer 2

如果您知道要选择的行号，那么使用漂亮汤中的select可以轻松完成此操作。（注意：这是bs4中的）

row = 5
while true
    element = soup.select('tr:nth-of-type('+ row +')')
    if len(element) > 0:
        # element is your desired row element, do what you want with it 
        row += 5
    else:
        break

Answer 3

作为一般解决方案，您可以将表转换为嵌套列表并迭代。。。

import BeautifulSoup

def listify(table):
  """Convert an html table to a nested list""" 
  result = []
  rows = table.findAll('tr')
  for row in rows:
    result.append([])
    cols = row.findAll('td')
    for col in cols:
      strings = [_string.encode('utf8') for _string in col.findAll(text=True)]
      text = ''.join(strings)
      result[-1].append(text)
  return result

if __name__=="__main__":
    """Build a small table with one column and ten rows, then parse into a list"""
    htstring = """<table> <tr> <td>foo1</td> </tr> <tr> <td>foo2</td> </tr> <tr> <td>foo3</td> </tr> <tr> <td>foo4</td> </tr> <tr> <td>foo5</td> </tr>  <tr> <td>foo6</td> </tr>  <tr> <td>foo7</td> </tr>  <tr> <td>foo8</td> </tr>  <tr> <td>foo9</td> </tr>  <tr> <td>foo10</td> </tr></table>"""
    soup = BeautifulSoup.BeautifulSoup(htstring)
    for idx, ii in enumerate(listify(soup)):
        if ((idx+1)%5>0):
            continue
        print ii

运行。。。

[mpenning@Bucksnort ~]$ python testme.py
['foo5']
['foo10']
[mpenning@Bucksnort ~]$