如何修复此属性错误python？

Question

我的代码有问题。我正在尝试提取本网站 (https://www.local.ch/en/q/geneve/employment%20agency?slot=yellow) 上列出的职位，其中包含公司名称及其信息的链接。第一部分有效，我可以打印所有名称，但是打印到其信息的链接会给我错误：

Traceback (most recent call last):
  File "main.py", line 20, in <module>
    href = (links.get("href"))
  File "/opt/virtualenvs/python3/lib/python3.8/site-packages/bs4/element.py", line 921, in __getattr__
    raise AttributeError(
AttributeError: 'NavigableString' object has no attribute 'get'

这是我的代码：

print("Hello, welcome to local job in geneva finder")

import requests
from bs4 import BeautifulSoup

url = "https://www.local.ch/en/q/geneve/employment%20agency?slot=yellow"

response = requests.get(url)
html = response.text

soup = BeautifulSoup(html, "html.parser")
names = soup.findAll("h2")
for name in names:
    print(name.text)

link = soup.find("a")
for links in link:
    href = (links.get("href"))
    if href.startswith("https://www.local.ch/en/d/geneve/1204/recruiting"):
        print(href)

Answer 1

使用findAll提取所有<a>标签。

links = soup.findAll("a")

迭代链接而不是名称的循环以从所有 <a> 标记中获取 href。
link.get("href") 可以在<a> 标签中找不到 href 的情况下也返回 None。所以写一个条件来检查天气是否为None。

完整代码：

print("Hello, welcome to local job in geneva finder")
import requests
from bs4 import BeautifulSoup

url = "https://www.local.ch/en/q/geneve/employment%20agency?slot=yellow"

response = requests.get(url)
html = response.text

soup = BeautifulSoup(html, "html.parser")
names = soup.findAll("h2")
for name in names:
    print(name.text)

links = soup.findAll("a")
for link in links:
    href = link.get("href")
    if href:
        if href.startswith("https://www.local.ch/en/d/geneve/1204/recruiting"):
            print(href)