从 <p> 中提取特定的标签标签答案

【问题标题】：Extract specific tag tag from the <p>从 <p> 中提取特定的标签标签
【发布时间】：2021-11-26 11:09:08
【问题描述】：

我只想从p 标签中提取address，比如我想得到这些Santa Barbara, CA 93101

[<p class="hide" id="phoneDiv_80863"><i aria-hidden="true" class="fa fa-phone-square"></i> (805) 636-9890</p>, <p>

Santa Barbara, CA 93101



</p>, <p style="margin-top:2em;"><a class="btn btn-default" href="/profile/id/80863/NicoleABotaitis93101" target="_top">View</a> <a class="btn btn-default" href="mailto:nicole@santabarbaratherapist.com" id="eml80863" target="_top">Email</a></p>]
[]
[<p class="hide" id="phoneDiv_26092"><i aria-hidden="true" class="fa fa-phone-square"></i> 8058956960</p>, <p>

Santa Barbara, CA 93111

代码

import requests
from bs4 import BeautifulSoup
import pandas as pd
import re

limit = 25

url = f'https://www.counselingcalifornia.com/cc/cgi-bin/utilities.dll/customlist?FIRSTNAME=~&LASTNAME=~&ZIP=&DONORCLASSSTT=&_MULTIPLE_INSURANCE=&HASPHOTOFLG=&_MULTIPLE_EMPHASIS=&ETHNIC=&_MULTIPLE_LANGUAGE=ENG&QNAME=THERAPISTLIST&WMT=NONE&WNR=NONE&WHP=therapistHeader.htm&WBP=therapistList.htm&RANGE=1%2F{limit}&SORT=LASTNAME'
headers = {'User-Agent': 'Mozilla/5.0 (Linux; Android 6.0; Nexus 5 Build/MRA58N) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.45 Mobile Safari/537.36'}

response = requests.get(url, headers=headers)
soup = BeautifulSoup(response.text, 'html.parser')
rows = soup.find_all('div', {'class':'row'})
temp=[]
for row in rows:
    t=row.find_all('div',class_='col-sm-3')
    for i in t:
        u=i.find_all('p')
        print(u)

【问题讨论】：

“地址”p 标签与“常规”p 标签的区别是什么？你怎么知道地址是什么？

标签： python web-scraping beautifulsoup

【解决方案1】：

这是使用css-selector 的更好解决方案：

import requests
from bs4 import BeautifulSoup
import pandas as pd

limit = 25

url = f'https://www.counselingcalifornia.com/cc/cgi-bin/utilities.dll/customlist?FIRSTNAME=~&LASTNAME=~&ZIP=&DONORCLASSSTT=&_MULTIPLE_INSURANCE=&HASPHOTOFLG=&_MULTIPLE_EMPHASIS=&ETHNIC=&_MULTIPLE_LANGUAGE=ENG&QNAME=THERAPISTLIST&WMT=NONE&WNR=NONE&WHP=therapistHeader.htm&WBP=therapistList.htm&RANGE=1%2F{limit}&SORT=LASTNAME'
headers = {'User-Agent': 'Mozilla/5.0 (Linux; Android 6.0; Nexus 5 Build/MRA58N) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.45 Mobile Safari/537.36'}

response = requests.get(url, headers=headers)
soup = BeautifulSoup(response.text, 'html.parser')

for address in soup.select('.col-sm-3>p:nth-child(3)'):
    print(address.text.strip())

样本输出：

Santa Barbara, CA 93101
Santa Barbara, CA 93111
Santa Barbara, CA 93101
Tustin, CA 92780
Valencia, CA 91355
Pasadena, CA 91105
United States
Walnut Creek, CA 94596
Woodland Hills, CA 91365-0644
Monterey, CA 93940
Granada Hills, CA 91344
United States
Studio City, CA 91604
Santa Rosa, CA 95404
Sonoma
San Dimas, CA 91773
United States
San Francisco, CA 94116
Rancho Mirage, CA 92270
Berkeley, CA 94705-1808
Anderson, CA 96007
Shasta
Mission Viejo, CA 92691
United States
Claremont, CA 91711
Seal Beach, CA 90740
USA
West Covina, CA 91790
Los Angeles
Mission Viejo, CA 92692
Laguna Niguel, CA 92677
Camarillo, CA 93010
West Hills, CA 91308

参考资料：

【讨论】：

【解决方案2】：

这是你要找的东西吗：

soup = BeautifulSoup(response.text, 'html.parser')
rows = soup.find_all('div', {'class':'row'})
temp=[]
for row in rows:
    t=row.find_all('div',class_='col-sm-3')
    for i in t:
        u=i.find_all('p')[1:2]
        for each_u in u:
            address = each_u.text.split('\n')[1]
            print(address)

【讨论】：

你能告诉我这些是什么意思吗[1:2]
@AmenAziz 请检查我发布的答案。
好的，我检查一下
我们必须对列表进行切片，而不仅仅是获取元素
@RupalShah 使用 3 个循环是完全没有必要的，并且会影响性能的解决方案