Question

我正在使用请求，Beautifulsoup4和django。结果工作正常，除了它们在括号中是封闭的。我如何在没有括号的情况下返回它们。继承我的代码：

google = requests.get(url)
bs = BeautifulSoup(google.content, 'html.parser')
d = bs.title.string

a_links = bs.findAll('div', 'entry')

links = []
for link in a_links:
    links.append((
        link.text,

    )
)

# samples = soup.find_all("a", "item-title")
myset = set(links)
context = {
    "links": links,
}
return render(request, 'index.html', context)

我的模板

{% for l in links %}
   {{l}}<br><hr>
{% endfor %}

我的结果

('251Drake Dethrones 50 Cent on Forbes Top Five Richest Hip-Hop Stars of 20168,787 views9 Hrs Ago',)

('175Kanye Claims Kim Kardashian Broke Boundaries for Biracial Couples5,923 views15 Hrs Ago',)

('94Stevie Wonder: If Michael Jackson is King of Pop, Prince is the Emperor4,482 views16 Hrs Ago',)

我想像这样显示它

'251Drake Dethrones 50 Cent on Forbes Top Five Richest Hip-Hop Stars of 20168,787 views9 Hrs Ago',

'175Kanye Claims Kim Kardashian Broke Boundaries for Biracial Couples5,923 views15 Hrs Ago',

'94Stevie Wonder: If Michael Jackson is King of Pop, Prince is the Emperor4,482 views16 Hrs Ago',

如何做到这一点

Answer 1

那是因为你实际上是将元组附加到列表中。替换：

links = []
for link in a_links:
    links.append((
        link.text,
    )
)

使用：

links = [link.text for link in a_links]

Answer 2

或者，如果您希望输出字面上（如有问题，包括单引号和尾随逗号），则可以跳过第一个和最后一个字符以排除括号：

links = [str((link.text,))[1:-1] for link in a_links]

如何从bs4结果

2 个答案: