用python爬取前程无忧招聘网

最新推荐文章于 2025-06-16 17:46:02 发布

原创最新推荐文章于 2025-06-16 17:46:02 发布 · 3.1k 阅读

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本文介绍了一种使用Python爬虫抓取前程无忧网站招聘信息的方法，通过requests库实现网页数据的获取，并解析出岗位名称、详情链接等关键信息。文章提供了完整的代码示例，适用于数据分析和招聘市场研究。

直接上代码了，相比前篇文章智联招聘网的数据，前程无忧网的数据可以爬取很多。

网址：https://search.51job.com/list/040000,000000,0000,00,9,99,%20,2,1.html?lang=c&stype=&postchannel=0000&workyear=99&cotype=99&degreefrom=99&jobterm=99&companysize=99&providesalary=99&lonlat=0%2C0&radius=-1&ord_field=0&confirmdate=9&fromType=&dibiaoid=0&address=&line=&specialarea=00&from=&welfare=
简化后：https://search.51job.com/list/040000,000000,0000,00,9,99,%20,2,1.html?
相比之下，这份获取的数据更适合练习学习数据分析。
爬取的方法跟步骤跟智联招聘网那篇一样。都是用到了第三方库requests

import requests
import re
import os
import time


class Spider(object):
    page_count = 0

    def __init__(self):
        self.headers = {
            'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/73.0.3683.103 Safari/537.36'}
        global path, page_count
        path = './前程无忧招聘网/'
        if not os.path.exists(path):
            os.mkdir(path)
        self.path = path
        with open(path + "前程无忧招聘.json", "a", encoding='utf-8') as fp:
            fp.write('{')

    def response(self, url, headers):
        """
        请求访问服务器获取资源
        """
        try:
            response = requests.get(url, headers)
            response.encoding = response.apparent_encoding
            return response
        except:
            print('访问失败')
            return None

    def parse(self, response):
        """
        解析出我们需要进入的岗位名称及其详情链接
        """
        res = re.findall(r'<span>\s+<a target="_blank" title="(.*?)" href="(.*?)" .*?</a>', response.text,
                         re.S)
        return res

    def parse_S(self, response):
        try:
            price = re.findall('<div class="cn".*?<strong>(.*?)</strong>', response.text, re.S)[0]
        except:
            price = None
        try:
            company_name = re.findall('class="cname">.*?title="(.*?)" class.*?<em.*?', response.text, re.S)[0]
        except:
            company_name = None
        try:
            add = re.findall('<p class="msg.*?title="(.*?)&nbsp;', response.text, re.S)[0]
        except:
            add = None
        try:
            num = re.findall('<p class="msg ltype".*?招(\d+)人&nbsp;', response.text, re.S)[0]
        except:
            num = '若干人'
        detail = {}
        detail['公司名称'] = company_name
        detail['工作城市'] = add
        detail['招聘人数'] = num
        detail['工资情况'] = price
        return detail

    def getContent(self, url):
        response = self.response(url, self.headers)
        detail = self.parse_S(response)
        return detail

    def save(self, items):
        Spider.page_count += 1
        print(Spider.page_count)
        with open(path + "前程无忧招聘.json", "a", encoding='utf-8') as fp:
            fp.write(str(items) + ',')

    def crawl(self, page):
        crawl_url = 'https://search.51job.com/list/040000,000000,0000,00,9,99,%2520,2,{}.html?'.format(page)
        response = self.response(crawl_url, self.headers)
        res = self.parse(response)
        items = {}
        for name, url in res:
            detail = self.getContent(url)
            items['name'] = name
            items['detail'] = detail
            self.save(items)

    def main(self):
        for page in range(0, 100):
            self.crawl(page)
        with open(path + "前程无忧招聘.json", "a", encoding='utf-8') as fp:
            fp.write('}')


start = time.time()
if __name__ == '__main__':
    c = Spider()
    c.main()
    end = time.time()
    print(end - start)

如发现可以改进的地方或者哪里做得不好，希望大家能够提出多多交流。