BeautifulSoup/Python - 将HTML表转换为CSV并获取一列的href

Question

我正在使用以下代码获取HTML表:

import csv
import urllib2
from bs4 import BeautifulSoup

with open('listing.csv', 'wb') as f:
    writer = csv.writer(f)
    for i in range(39):
        url = "file:///C:/projects/HTML/Export.htm".format(i)
        u = urllib2.urlopen(url)
        try:
            html = u.read()
        finally:
            u.close()
        soup=BeautifulSoup(html)
        for tr in soup.find_all('tr')[2:]:
            tds = tr.find_all('td')
            row = [elem.text.encode('utf-8') for elem in tds]
            writer.writerow(row)

一切都很完美,但我试图抓住第9列Href URL.它目前给我txt值但不是URL.

另外,我的HTML中有两个表,无论如何要跳过第一个表并使用第二个表构建csv文件？

任何帮助都非常受欢迎,因为我是Python的新手,需要这个项目我自动进行每日转换.

非常感谢!

Answer 1

您应该访问第 8 个标签内href的标签属性：atd

import csv
import urllib2
from bs4 import BeautifulSoup

records = []
for index in range(39):
    url = get_url(index)  # where is the formatting in your example happening?
    response = urllib2.urlopen(url)
    try:
        html = response.read()
    except Exception:
        raise
    else:
        my_parse(html)
    finally:
        try:
            response.close()
        except (UnboundLocalError, NameError):
            raise UnboundLocalError

def my_parse(html):
    soup = BeautifulSoup(html)
    table2 = soup.find_all('table')[1]
    for tr in table2.find_all('tr')[2:]:
        tds = tr.find_all('td')
        url = tds[8].a.get('href')
        records.append([elem.text.encode('utf-8') for elem in tds])
        # perhaps you want to update one of the elements of this last
        # record with the found url now?

# It's more efficient to write only once
with open('listing.csv', 'wb') as f:
    writer = csv.writer(f)
    writer.writerows(records)

我冒昧地定义了一个get_url基于索引的函数，因为您的示例每次都会重新读取同一个文件，这是我猜您实际上并不想要的。我将把实施留给你。另外，我还添加了一些更好的异常处理。

同时，我展示了如何从该网页的表格访问第二个表格。