动态查找href标记

Question

我试图从我的靓汤搜索中提取“信息技术”作为输出。但我还没弄清楚，因为“扇区”是URL中任何一种股票代码的动态值。你知道吗

有人能告诉我如何提取这些信息吗？你知道吗

<a href="http://eresearch.fidelity.com/eresearch/markets_sectors/sectors/sectors_in_market.jhtml?tab=learn&amp;sector=45">Information Technology</a>

我的代码：

url = 'https://eresearch.fidelity.com/eresearch/goto/evaluate/snapshot.jhtml?symbols=AAPL'

html = requests.get(url).text    
detail_tags_sector = BeautifulSoup(html, 'lxml')
detail_tags_sector.find_all('a')

Answer 1

要从锚元素获取文本，需要访问每个锚元素上的.text变量
因此，您的代码将更改为：

url = 'https://eresearch.fidelity.com/eresearch/goto/evaluate/snapshot.jhtml?symbols=AAPL'
contents = []

html = requests.get(url).text    
detail_tags_sector = BeautifulSoup(html, 'html.paser')
for anchor in detail_tags_sector.find_all('a'):
    contents.append(anchor.text)
print(contents)

Answer 2

这些答案的问题在于，它们收集了页面上所有链接的文本，并且有相当多的链接。如果只选择information technology字符串，则只需添加：

info = soup.select_one('[href*="sectors_in"]')
print(info.text)

输出：

Information Technology

Answer 3

您可以使用以下任一选项。你知道吗

import requests
from lxml.html.soupparser import fromstring
url = 'https://eresearch.fidelity.com/eresearch/goto/evaluate/snapshot.jhtml?symbols=AAPL'
html = requests.get(url).text
soup=fromstring(html)
findSearch = soup.xpath('//a[contains(text(), "Information Technology")]/text()')
print(findSearch[0])

或者

from bs4 import BeautifulSoup
from lxml import html
import requests
url = 'https://eresearch.fidelity.com/eresearch/goto/evaluate/snapshot.jhtml?symbols=AAPL'

html = requests.get(url).text
detail_tags_sector = BeautifulSoup(html, 'lxml')
for link in detail_tags_sector.find_all('a'):
    print(link.text)

或者

from bs4 import BeautifulSoup    
import requests
url = 'https://eresearch.fidelity.com/eresearch/goto/evaluate/snapshot.jhtml?symbols=AAPL'
html = requests.get(url).text
soup = BeautifulSoup(html, 'html.parser')
for link in soup.find_all('a'):
    print(link.text)

如果有帮助请告诉我。你知道吗