问如何在python中从img html中抓取src
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Stack Overflow用户

提问于 2019-08-23 11:06:02

回答 2查看 525关注 0票数 1

我试图抓取img的src，但我找到的代码返回了许多img src，但不是我想要的那个。我不知道我做错了什么。我在"https://www.tripadvisor.dk/Restaurant_Review-g189541-d15804886-Reviews-The_Pescatarian-Copenhagen_Zealand.html“上抓取TripAdvisor

这就是我要提取的HTML代码片段：

 <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

我试过了：

    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'))

返回的内容与此类似，但不是我需要的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==

python

image

web-scraping

回答 2

Stack Overflow用户

回答已采纳

发布于 2019-08-23 16:13:31

您可以使用requests和re来做到这一点。它只是src的坐标部分，它是基于位置的变量。

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)

票数 1

Stack Overflow用户

发布于 2019-08-23 11:33:42

Selenium是一种变通方法，我对它进行了测试，效果非常好。您的位置如下：

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)