如何访问此网站的下一页



有一个网站,我想从中提取特定的链接。我已经设法做到了,但只针对一个网站。有133个网站,我需要从中链接。你能告诉我怎么做吗?

到目前为止,我能够建造这个。我理解它应该以某种方式使用数组";页面";但我不知道如何告诉脚本循环使用它,并将其视为新站点。提前谢谢。

from bs4 import BeautifulSoup
import urllib.request
import pandas as pd
import requests
import time
pages = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133]
links = []
for page in pages:
url =  urllib.request.urlopen("https://www.derekprince.org/Media/")
content = url.read()
soup = BeautifulSoup(content)
result = soup.findAll('a', {"class": "media_recording_file_download" })
links.append(result)

再次,Selenium成为web抓取问题的最简单、最苛刻的解决方案:(如果有人需要它或有类似的问题,下面是解决方案。我已经使用googlechrome来复制xpath并查找类名。

from selenium import webdriver                                                                            
pages = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133]

driver = webdriver.Chrome("/home/grzegorz/Documents/chromedriver")                                        
driver.get("https://www.derekprince.org/Media/")                                                          
driver.find_elements_by_class_name("media_recording_file_download")
for i in pages:
driver.find_element_by_xpath("//*[@id='media_pager_top']/a[2]").click()         
for i in driver.find_elements_by_class_name("media_recording_file_download"): 
i.click()

一个问题有多个问题-所以我建议您改进它。

我只是在详细回答第一个问题,请再问一个新问题。

如何处理迭代

代替list,您可以使用range()-将两个参数(start,stop(传递给range(),它将生成从起始数字到stop-1的整数。

for i in range(1,6):
print(f'This is my iteration #{i}')

要将您的变量和string连接起来,您可以使用pythons f'string。

提示你的下一个问题

此网站处理form,因此您必须执行包含page变量的发布请求。

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