Python 无法得到美味的汤
我正在尝试获取锚点(a)元素的超链接,但我不断获得: https://in.finance.yahoo.com/h ttps://in.finance.yahoo.com/ 我已经尝试了这里提供的所有解决方案: 这是我的密码:Python 无法得到美味的汤,python,html,beautifulsoup,Python,Html,Beautifulsoup,我正在尝试获取锚点(a)元素的超链接,但我不断获得: https://in.finance.yahoo.com/h ttps://in.finance.yahoo.com/ 我已经尝试了这里提供的所有解决方案: 这是我的密码: href_links = [] symbols = [] prices = [] commodities = [] CommoditiesUrl = "https://in.finance.yahoo.com/commodities" r = requests.get
href_links = []
symbols = []
prices = []
commodities = []
CommoditiesUrl = "https://in.finance.yahoo.com/commodities"
r = requests.get(CommoditiesUrl)
data = r.text
soup = BeautifulSoup(data)
counter = 40
for i in range(40, 404, 14):
for row in soup.find_all('tbody'):
for srow in row.find_all('tr'):
for symbol in srow.find_all('td', attrs={'class':'data-col0'}):
symbols.append(symbol.text)
href_link = soup.find('a').get('href')
href_links.append('https://in.finance.yahoo.com/' + href_link)
for commodity in srow.find_all('td', attrs={'class':'data-col1'}):
commodities.append(commodity.text)
for price in srow.find_all('td', attrs={'class':'data-col2'}):
prices.append(price.text)
pd.DataFrame({"Links": href_links, "Symbol": symbols, "Commodity": commodities, "Prices": prices })
此外,我想知道是否可行,类似于网站,在我的熊猫数据框中将商品的符号作为超链接
我不确定你发布的代码是怎么回事,但是你可以通过
查找一个属性为数据符号设置为GC=F
的元素来获取该URL。html有两个这样的元素。您想要的是第一个,它是由soup.find('a',{'data-symbol':'GC=F'})返回的。get('href')
此外,我想知道是否可行,类似于网站,在我的熊猫数据框中将商品的符号作为超链接
我不熟悉熊猫,但我会说答案是肯定的。请参阅:最外层的for
环路到底是用来做什么的<范围(4040414)中的i的code>:循环体中甚至没有引用i
i
import requests, urllib
from bs4 import BeautifulSoup
CommoditiesUrl = "https://in.finance.yahoo.com/commodities"
r = requests.get(CommoditiesUrl)
data = r.text
soup = BeautifulSoup(data)
gold_href = soup.find('a', {'data-symbol': 'GC=F'}).get('href')
# If it is a relative URL, we need to transform it into an absolute URL (it always is, fwiw)
if not gold_href.startswith('http'):
# If you insist, you can do 'https://in.finance.yahoo.com" + gold_href
gold_href = urllib.parse.urljoin(CommoditiesUrl, gold_href)
print(gold_url)