Python BeautifulSoup刮表

Question

我正在尝试用BeautifulSoup创建一个表刮。我写了这个Python代码：

import urllib2
from bs4 import BeautifulSoup

url = "http://dofollow.netsons.org/table1.htm"  # change to whatever your url is

page = urllib2.urlopen(url).read()
soup = BeautifulSoup(page)

for i in soup.find_all('form'):
    print i.attrs['class']

我需要刮Nome，Cognome，Email。

Answer 1

循环表格行（tr标记）并获取单元格文本（td标记）：

for tr in soup.find_all('tr')[2:]:
    tds = tr.find_all('td')
    print "Nome: %s, Cognome: %s, Email: %s" % \
          (tds[0].text, tds[1].text, tds[2].text)

打印：

Nome:  Massimo, Cognome:  Allegri, Email:  Allegri.Massimo@alitalia.it
Nome:  Alessandra, Cognome:  Anastasia, Email:  Anastasia.Alessandra@alitalia.it
...

仅供参考，这里的[2:]切片是跳过两个标题行。

UPD，这里是如何将结果保存到txt文件中的：

with open('output.txt', 'w') as f:
    for tr in soup.find_all('tr')[2:]:
        tds = tr.find_all('td')
        f.write("Nome: %s, Cognome: %s, Email: %s\n" % \
              (tds[0].text, tds[1].text, tds[2].text))

Answer 2

# Libray
from bs4 import BeautifulSoup

# Empty List
tabs = []

# File handling
with open('/home/rakesh/showHW/content.html', 'r') as fp:
    html_content = fp.read()

    table_doc = BeautifulSoup(html_content, 'html.parser')
    # parsing html content
    for tr in table_doc.table.find_all('tr'):
        tabs.append({
            'Nome': tr.find_all('td')[0].string,
            'Cogname': tr.find_all('td')[1].string,
            'Email': tr.find_all('td')[2].string
            })

    print(tabs)

Answer 3

OP发布的原始链接已死...但是您可以通过以下方式用gazpacho抓取表格数据：

第1步-导入Soup并下载html：

from gazpacho import Soup

url = "https://en.wikipedia.org/wiki/List_of_multiple_Olympic_gold_medalists"
soup = Soup.get(url)

第2步-查找表和表行：

table = soup.find("table", {"class": "wikitable sortable"}, mode="first")
trs = table.find("tr")[1:]

第3步-使用提取所需数据的函数来解析每一行：

def parse_tr(tr):
    return {
        "name": tr.find("td")[0].text,
        "country": tr.find("td")[1].text,
        "medals": int(tr.find("td")[-1].text)
    }

data = [parse_tr(tr) for tr in trs]
sorted(data, key=lambda x: x["medals"], reverse=True)