python - 将 href 字符串转换为链接列表

Question

我正在尝试使用以下代码从 Gosugamers 抓取一些统计数据，包括比赛结果和这些比赛的球队名称:

from bs4 import BeautifulSoup
import requests

for i in range(411):
    try:
        i += 1
        print(i)
        url = 'http://www.gosugamers.net/counterstrike/gosubet?r-page={}'.format(i)
        r = requests.get(url)
        web = BeautifulSoup(r.content,"html.parser")
        table = web.findAll("table", attrs={"class":"simple matches"})
        table = table[1]
        links = table('a')
        for link in links:
            if 'matches' in link.get('href', None):
                if len(link.get('href', None)) != 0:
                    print(link.get('href', None))

    except:
        pass

但是当我得到 link.get('href', None) 时，它是一个包含所有链接的字符串，用于单个页面上的匹配，我不知道如何把它变成所有链接的列表，如果有人能帮助我，我会很高兴，谢谢!

Answer 1

在我看来，link.get('href', None) 实际上返回了一个链接。 get 方法文档说:

bs4.element.Tag实例的get(self, key, default=None)方法

Returns the value of the 'key' attribute for the tag, or
the value given for 'default' if it doesn't have that
attribute.

因此，当您获得一个包含“匹配项”的链接时，您可以将其添加到列表中。

from bs4 import BeautifulSoup
import requests

all_links = []

i = 1
for i in range(411):
    try:
        print(i)
        url = 'http://www.gosugamers.net/counterstrike/gosubet?r-page={}'.format(i)
        r = requests.get(url)
        web = BeautifulSoup(r.content,"html.parser")
        table = web.findAll("table", attrs={"class":"simple matches"})
        table = table[1]
        links = table('a')

        for link in links:
            href = link.get('href')
            if href is not None and 'matches' in href:
                all_links.append(href)

        i += 1
    except:
        pass

print "Here are all the links: ", all_links