问题描述
我正在尝试使用BeautifulSoup提取链接的标题。我正在使用的代码如下:
import requests
from bs4 import BeautifulSoup
from urllib.parse import urljoin
import pandas as pd
hdr={'User-Agent':'Chrome/84.0.4147.135'}
frame=[]
for page_number in range(19):
http= "https://www.epa.wa.gov.au/media-statements?page={}".format(page_number+1)
print('Downloading page %s...' % http)
url= requests.get(http,headers=hdr)
soup = BeautifulSoup(url.content,'html.parser')
for row in soup.select('.view-content .views-row'):
content = row.select_one('.views-field-body').get_text(strip=True)
title = row.text.strip(':')
link = 'https://www.epa.wa.gov.au' + row.a['href']
date = row.select_one('.date-display-single').get_text(strip=True)
frame.append({
'title': title,'link': link,'date': date,'content': content
})
dfs = pd.DataFrame(frame)
dfs.to_csv('epa_scrapper.csv',index=False,encoding='utf-8-sig')
但是,运行上述代码后,什么都没有显示。如何提取存储在链接中的定位标记的title属性内的值?
此外,我只想知道如何将“标题”,“链接”,“ dt”,“内容”附加到csv文件中。
非常感谢您。
解决方法
要获取链接文本,可以使用选择器"h5 a"
。例如:
import requests
from bs4 import BeautifulSoup
from urllib.parse import urljoin
import pandas as pd
hdr={'User-Agent':'Chrome/84.0.4147.135'}
frame=[]
for page_number in range(1,20):
http= "https://www.epa.wa.gov.au/media-statements?page={}".format(page_number)
print('Downloading page %s...' % http)
url= requests.get(http,headers=hdr)
soup = BeautifulSoup(url.content,'html.parser')
for row in soup.select('.view-content .views-row'):
content = row.select_one('.views-field-body').get_text(strip=True,separator='\n')
title = row.select_one('h5 a').get_text(strip=True)
link = 'https://www.epa.wa.gov.au' + row.a['href']
date = row.select_one('.date-display-single').get_text(strip=True)
frame.append({
'title': title,'link': link,'date': date,'content': content
})
dfs = pd.DataFrame(frame)
dfs.to_csv('epa_scrapper.csv',index=False,encoding='utf-8-sig')
创建epa_scrapper.csv
(来自LibreOffice的屏幕截图):