Question

我正在尝试抓取img的src，但是我发现的代码返回了许多img src，但不是我想要的那个。我不知道自己在做什么错。我正在“ https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html”上刮擦TripAdvisor

这是我要从中提取的HTML代码段：

 <div class="restaurants-detail-overview-cards-LocationOverviewCard__cardColumn--2ALwF"><h6>Placering og kontaktoplysninger</h6><span><div><span data-test-target="staticMapSnapshot" class=""><img class="restaurants-detail-overview-cards-LocationOverviewCard__mapImage--22-Al" src="https://trip-raster.citymaps.io/staticmap?scale=1&amp;zoom=15&amp;size=347x137&amp;language=da&amp;center=55.687988,12.596316&amp;markers=icon:http%3A%2F%2Fc1.tacdn.com%2F%2Fimg2%2Fmaps%2Ficons%2Fcomponent_map_pins_v1%2FR_Pin_Small.png|55.68799,12.596316"></span></div></span>

我希望代码返回：（来自src的子字符串）

55.68799,12.596316

我尝试过：

    import pandas as pd
    pd.options.display.max_colwidth = 200
    from urllib.request import urlopen
    from bs4 import BeautifulSoup as bs
    import re

    web_url = "https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html"
    url = urlopen(web_url)
    url_html = url.read()

    soup = bs(url_html, 'lxml')
    soup.find_all('img')

    for link in soup.find_all('img'):
        print(link.get('src'))

返回的是此行的内容，而不是我需要的src：

https://static.tacdn.com/img2/branding/rebrand/TA_logo_secondary.svg
https://static.tacdn.com/img2/branding/rebrand/TA_logo_primary.svg 
https://static.tacdn.com/img2/branding/rebrand/TA_logo_secondary.svg
data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==
data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==

Answer 1

Selenium是一种变通方法，我对其进行了测试，并且看起来很有魅力。您在这里：

\b\d{6,17}\b

上述结果 _

如果您以前没有使用过from selenium import webdriver driver = webdriver.Chrome('chromedriver.exe') driver.get("https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html") links = driver.find_elements_by_xpath("//*[@src]") urls = [] for link in links: url = link.get_attribute('src') if '|' in url: urls.append(url.split('|')[1]) # saves in a list only the numbers you want i.e. 55.68799,12.596316 print(url) print(urls)，则可以找到网络驱动程序https://chromedriver.storage.googleapis.com/index.html?path=2.46/

或这里

https://sites.google.com/a/chromium.org/chromedriver/downloads

Answer 2

您可以仅执行请求并重新执行此操作。只是src的坐标部分是基于位置的变量。

import requests, re

p = re.compile(r'"coords":"(.*?)"')
r = requests.get('https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html')
coords = p.findall(r.text)[1]
src = f'https://trip-raster.citymaps.io/staticmap?scale=1&zoom=15&size=347x137&language=da&center={coords}&markers=icon:http://c1.tacdn.com//img2/maps/icons/component_map_pins_v1/R_Pin_Small.png|{coords}'
print(src)
print(coords)

如何从python中的img html抓取src

2 个答案: