"<i>beatiful soup 4 getting an output.<\/i>美丽的汤4得到输出。<\/b> <i>How to change as a required format?<\/i>如何更改为所需格式？<\/b> <i>([&#39;link1&#39;, &#39;link2&#39;, &#39;link3&#39;])<\/i> ([&#39;link1&#39;, &#39;link2&#39;, &#39;link3&#39;])<\/b>"

Question

美丽的汤4得到输出。 如何更改为所需格式？

我得到这个低于输出。

['link1']
['link2']
['link3']

我需要一个如下所述的输出来形成一个数据框，所以我现在需要做什么。

['link1', 'link2', 'link3']

用代码解释也很好。 请帮我解决这个问题，在此先感谢。

我的代码

import bs4
from bs4 import BeautifulSoup
from csv import writer
import requests
headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:96.0) Gecko/20100101 Firefox/96.0'}
HOST = 'https://www.zocdoc.com'
#PAGE = 'gastroenterologists/2'
web_page = 'https://www.zocdoc.com/search?address=Houston%2C%20TX&insurance_carrier=&city=Houston&date_searched_for=&day_filter=AnyDay&filters=%7B%7D&gender=-1&language=-1&latitude=29.7604267&locationType=placemark&longitude=-95.3698028&offset=1&insurance_plan=-1&reason_visit=386&search_query=Gastroenterologist&searchType=specialty&sees_children=false&after_5pm=false&before_10am=false&sort_type=Default&dr_specialty=106&state=TX&visitType=inPersonVisit&&timesgridType='
with requests.Session() as session:
    (r := session.get(HOST, headers=headers)).raise_for_status()
    #(r := session.get(f'{HOST}/{PAGE}', headers=headers)).raise_for_status()
    (r := session.get(f'{web_page}', headers=headers)).raise_for_status()
    # process content from here
print(r.text)
soup = BeautifulSoup(r.text, 'lxml')
soup
print(soup.prettify())

作为链接获取的代码 1

for item in soup.find_all('img'):
    images = []
    items = (item['src'])
    images = 'https:'+items
    print(images)

代码2得到下面提到的输出格式

for item in soup.find_all('img'):
    c = []
    items = (item['src'])
    image = ('https:'+items)
    c.append(image)
    print(c)

输出 - ['link1'] 。 . ['链接']

Answer 1

原因是您在每个 for 循环中设置一个列表，然后覆盖它。 首先定义列表，然后附加到它，如下所示。

images = []
for item in soup.find_all('img'):
    items = (item['src'])
    images = [f"https:{i}" for i in items]
    print(images)

Answer 2

您必须将 url 附加到循环外的列表中，以避免覆盖并获得您期望的结构：

images = []
for item in soup.find_all('img'):
    images.append('https:'+item['src'])