返回使用Beautifoulsoup的特定标题关键字

Question

我正在尝试创建一个网络刮板，仅当标题中的RSS feed（XML格式）中有某个关键字时，才能返回文章。但是，每当我运行代码时，它也会返回空白，即使文章的标题本身正确运行（例如，标题都会正确打印，但是当我要求它返回时，当标题中有“说”一词时，即使“说”一词实际上是标题：

示例

xml_text = requests.get('https://nypost.com/feed/').text
soup = BeautifulSoup(xml_text, 'xml')
ny_rss_search = soup.find_all("Mark")
ny_rss_title3 = soup.find_all('title')
ny_rss_url3 = soup.find_all('link')
ny_rss_summary3 = soup.find_all('description')
ny_rss_url_compact3 = ny_rss_url3[2].text.strip()
if 'Guide' in ny_rss_title3:
    webbrowser.open(ny_rss_url_compact3, new=2)
    print(f'NY Post Article Title: {ny_rss_title3[1].text.strip()}\n')
    print(f"NY Post Article URL: {ny_rss_url3[2].text.strip()}\n")
    print(f'NY Post Article Summary: {ny_rss_summary3[1].text.strip()}\n')
    winsound.PlaySound("notify.wav", winsound.SND_ALIAS)

html文本：

<?xml version="1.0" encoding="UTF-8"?>
<rss version="2.0"
    xmlns:content="http://purl.org/rss/1.0/modules/content/"
    xmlns:dc="http://purl.org/dc/elements/1.1/"
    xmlns:atom="http://www.w3.org/2005/Atom"
    xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
    xmlns:slash="http://purl.org/rss/1.0/modules/slash/"
        
    xmlns:georss="http://www.georss.org/georss"
    xmlns:geo="http://www.w3.org/2003/01/geo/wgs84_pos#"
    xmlns:media="http://search.yahoo.com/mrss/"
        >

<channel>
    <title>New York Post</title>
    <atom:link href="https://nypost.com/feed/" rel="self" type="application/rss+xml" />
    <link>https://nypost.com</link>
    <description>Your source for breaking news, news about New York, sports, business, entertainment, opinion, real estate, culture, fashion, and more.</description>
    <lastBuildDate>Tue, 05 Jul 2022 14:06:44 +0000</lastBuildDate>
    <language>en-US</language>
    <sy:updatePeriod>hourly</sy:updatePeriod>
    <sy:updateFrequency>1</sy:updateFrequency>
    <generator>https://wordpress.org/?v=5.9.3</generator>
            <item>
            <title>Blue Jays coach Mark Budzinski&#8217;s daughter Julia died in boating accident</title>
            <comments>https://nypost.com/2022/07/05/mark-budzinskis-daughter-julia-17-died-in-boating-accident/#respond</comments>
            <pubDate>Tue, 05 Jul 2022 10:01:06 -0400</pubDate>

                            <link>https://nypost.com/2022/07/05/mark-budzinskis-daughter-julia-17-died-in-boating-accident/</link>
                <dc:creator>Associated Press</dc:creator>
            
            <guid isPermaLink="false">https://nypost.com/?post_type=article&#038;p=22918233</guid>
                                                        <description><![CDATA[Pearson said no foul play is suspected and alcohol was not a factor. “It was a terrible accident,” she said.]]></description>
                <content:encoded><![CDATA[Pearson said no foul play is suspected and alcohol was not a factor. “It was a terrible accident,” she said.]]></content:encoded>
                                    <enclosure url="https://nypost.com/wp-content/uploads/sites/2/2022/07/Julia-Budzinski.jpg?quality=90&#038;strip=all" type="image/jpeg" />
                                    <slash:comments>0</slash:comments>
            
        <media:content url="https://nypost.com/wp-content/uploads/sites/2/2022/07/Julia-Budzinski.jpg?w=1024" medium="image">
            <media:title type="html">The Blue Jays held a moment of silence for first base coach Mark Budzinski&#039;s daughter Julia on Sunday.</media:title>
        </media:content>

        <media:content url="https://nypost.com/wp-content/uploads/sites/2/2022/07/Mark-Budzinski.jpg?w=1024" medium="image">
            <media:title type="html">Mark Budzinski</media:title>
        </media:content>

Answer 1

您必须在feed的项目上迭代，并检查标题文本包含您的术语：

for e in soup.select('item'):
    if 'Guide' in e.title.text:
        print(e.title.text)
        print(e.link.text)

示例

from bs4 import BeautifulSoup
import requests

xml_text = requests.get('https://nypost.com/feed/').text
soup = BeautifulSoup(xml_text, 'xml')

for e in soup.select('item'):
    if 'Guide' in e.title.text:
        print(e.title.text)
        print(e.link.text)