尝试使用soup.select 和soup.find_all 提取url

Question

这是网页 HTML 源代码的一部分：

<a href="http://www.abcde.com"> <img style="width:100%" src="/FileUploads/B/763846f.jpg" alt="search" title="search" /></a>
<a id="parts_img01" href="/Result?s=9&amp;type=%E4&amp;name=%E9"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>apple</h4></a>
<a id="parts_img02" href="/Result?s=12&amp;type=%E4&amp;name=%E4"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>banana</h4></a>
<a id="parts_img03" href="/Result?s=10&amp;type=%E4&amp;name=%E8"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>cherry</h4></a>
<a id="parts_img07" href="/Result?s=14&amp;type=%E4&amp;name=%E8"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>melon</h4></a>

我想提取我想要的网址，比如以 /Result 开头的网址？ 我刚刚了解到您可以在美丽的汤中使用soup.find_all 和soup.select。

soup.find_all：

icon = soup.find_all(id = re.compile("parts_img"))

其中一个结果将成功打印：

<a href="/Result?s=9&amp;type=%E4&amp;name=%E9" id="parts_img01"><h4 style=""><i aria-hidden="true" class="fa f-c"></i>apple</h4></a>

汤选择：

for item in soup.select(".fa f-c"):
    print(item['href'])

这行不通……

有没有一种方法可以直接从 html 中提取 url？ 我只想打印：

/Result?s=9&type=%E4&name=%E9
/Result?s=12&type=%E4&name=%E4
/Result?s=10&type=%E4&name=%E8
/Result?s=14&type=%E4&name=%E8

Answer 1

要在不使用正则表达式的情况下获得相同的输出：

html = """
 <a href="http://www.abcde.com"> <img style="width:100%" src="/FileUploads/B/763846f.jpg" alt="search" title="search" /></a>
<a id="parts_img01" href="/Result?s=9&amp;type=%E4&amp;name=%E9"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>apple</h4></a>
<a id="parts_img02" href="/Result?s=12&amp;type=%E4&amp;name=%E4"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>banana</h4></a>
<a id="parts_img03" href="/Result?s=10&amp;type=%E4&amp;name=%E8"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>cherry</h4></a>
<a id="parts_img07" href="/Result?s=14&amp;type=%E4&amp;name=%E8"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>melon</h4></a>
"""
from bs4 import BeautifulSoup

soup = BeautifulSoup(html, "lxml")
for link in soup.select("[id^='parts_img']"):
    print(link['href'])

结果：

/Result?s=9&type=%E4&name=%E9
/Result?s=12&type=%E4&name=%E4
/Result?s=10&type=%E4&name=%E8
/Result?s=14&type=%E4&name=%E8

Answer 2

我认为这段代码将说明从给定的 html 中提取 href。

 html = """<a href="http://www.abcde.com"> <img style="width:100%" src="/FileUploads/B/763846f.jpg" alt="search" title="search" /></a>
<a id="parts_img01" href="/Result?s=9&amp;type=%E4&amp;name=%E9"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>apple</h4></a>
<a id="parts_img02" href="/Result?s=12&amp;type=%E4&amp;name=%E4"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>banana</h4></a>
<a id="parts_img03" href="/Result?s=10&amp;type=%E4&amp;name=%E8"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>cherry</h4></a>
<a id="parts_img07" href="/Result?s=14&amp;type=%E4&amp;name=%E8"><h4 style=""><i class="fa f-c" aria-hidden="true"></i>melon</h4></a>"""
from bs4 import BeautifulSoup as Soup
import re
from urllib.parse import urljoin
parser = Soup(html, "lxml")
href = [ urljoin("http://www.abcde.com", a["href"]) for a in parser.findAll("a", {"id" : re.compile('parts_img.*')})]
print(href)

Answer 3

我正在使用

#!/usr/bin/python

import requests
from bs4 import BeautifulSoup
import re

top_url = 'https://a-certain.org/item-index'
response = requests.get(top_url)
html = response.content
soup = BeautifulSoup(html, 'html.parser')
items = soup.select('a[href^="http://a-certain.org/items"]')
for item in items:
        print(items['href'])

输出是

http://a-certain.org/items/item1/
http://a-certain.org/items/item2/
http://a-certain.org/items/item3/