我怎样才能从 'every page' 得到 'page links'？

Question

我想通过 python3 从 'every page' 得到 'every page links'。

'every page' 位置在我的代码中位于 BaseUrl 中。而且，每个页面链接都位于我的代码中。

其中，

BaseUrl = 'https://www.jobplanet.co.kr/companies?sort_by=review_compensation_cache&industry_id=700&page='

select body = #listCompanies > div > div.section_group > section:nth-child(1) > div > div > dl.content_col2_3.cominfo > dt > a'

请检查我的代码。我想收集每个页面的每个链接，以便将 Link 列表作为 linkUrl。有什么问题吗？

from bs4 import BeautifulSoup
import csv
import os
import re
import requests
import json

# jobplanet
BaseUrl = 'https://www.jobplanet.co.kr/companies?sort_by=review_compensation_cache&industry_id=700&page='


for i in range(1, 5, 1):
        url = BaseUrl + str(i)
        r = requests.get(url)
        soup = BeautifulSoup(r.text,'lxml')
        body = soup.select('#listCompanies > div > div.section_group > section:nth-child(1) > div > div > dl.content_col2_3.cominfo > dt > a')
        #print(body)

        linkUrl = []
        for item in body:
            link = item.get('href')
            linkUrl.append(link)
print(linkUrl)

Answer 1

您选择的 CSS 选择器 returns 只有一个 record.I 提供了更简单的 css 选择器 returns 每页所有 10 条记录。

您需要在循环外定义列表。

from bs4 import BeautifulSoup
import requests

linkUrl = []
BaseUrl = 'https://www.jobplanet.co.kr/companies?sort_by=review_compensation_cache&industry_id=700&page={}'
for i in range(1, 6):
    url = BaseUrl.format(i)
    r = requests.get(url)
    soup = BeautifulSoup(r.text,'lxml')
    links=soup.select(".us_titb_l3 >a")
    for item in links:
        link = item.get('href')
        linkUrl.append(link)

print(linkUrl)

Answer 2

你的 Css 选择器是错误的还添加了分页

from bs4 import BeautifulSoup
import csv
import os
import re
import requests
import json
from urllib import parse

# jobplanet
BaseUrl = 'https://www.jobplanet.co.kr/companies?sort_by=review_compensation_cache&industry_id=700&page={}'
source  =  requests.get(BaseUrl.format(1))
soup = BeautifulSoup(source.text,'lxml')
last_page_index = soup.select('a[class="btn_pglast"]') # getting the last page index 
last_page_index = int(last_page_index[0].get('href').split('page=')[1]) if last_page_index else 1
for i in range(1, last_page_index):
    print('## Getting Page {} out of {}'.format(i,last_page_index))
    if i > 1: # to avoid getting the same page again
        url = BaseUrl.format(i)
        r = requests.get(url)
        soup = BeautifulSoup(r.text,'lxml')
    body = soup.select('dt[class="us_titb_l3"] a')
    linkUrl = []
    for item in body:
        link = item.get('href')
        link = parse.urljoin(BaseUrl, link)
        linkUrl.append(link)
print(linkUrl)

我怎样才能从 'every page' 得到 'page links'？

How can I get 'page links' from 'every page'?

beautifulsoup

href

web-crawler

web-scraping

python-requests