如果找到某些字符串，则提取链接和文本

Question

我正在尝试运行beautifulSoup从网站中提取链接和文本（我有权限）

我运行以下代码来获取链接和文本：

import requests
from bs4 import BeautifulSoup 

url = "http://implementconsultinggroup.com/career/#/6257"
r = requests.get(url)

soup = BeautifulSoup(r.content)

links = soup.find_all("a")

for link in links:
     if "career" in link.get("href"):
             print "<a href='%s'>%s</a>" %(link.get("href"), link.text)

结果如下：

^{pr2}$

这几乎是正确的，但是我只希望在文本中有哥本哈根的名字的情况下返回这些立场（即上面的马尔默立场不应该被返回）。在

站点的HTML代码如下所示：

<div class="small-12 medium-9 columns top-lined">
                                    <a href="/career/management-consultants-within-supply-chain-management/" class="box-link">
                                    <h2 class="article__title--tiny" data-searchable-text="">Management consultants within supply chain management</h2>
                                    <p class="article__longDescription" data-searchable-text="">COPENHAGEN • We are looking for bright graduates with a passion for supply chain management and supply chain planning for our planning and execution excellence team.</p>
                                    <div class="styled-link styled-icon">
                                        <span class="icon icon-icon">
                                            <i class="fa fa-chevron-right"></i>
                                        </span>
                                        <span class="icon-text">View Position</span>
                                    </div>
                                </a>
                            </div>

Answer 1

似乎您可以添加另一个条件：

(...)
for link in links:
    if "career" in link.get("href") and 'COPENHAGEN' in link.text:
       print "<a href='%s'>%s</a>" %(link.get("href"), link.text)