如何创建文件并将抓取的数据保存在其中？

Question

我已经编写了这个脚本，但我尝试了几个选项来保存数据，但我一直在弄乱代码。如何将提取的数据保存到 csv 或 excel 文件中？

import requests
from bs4 import BeautifulSoup

base_url = "http://www.privredni-imenik.com/pretraga?abcd=&keyword=&cities_id=0&category_id=0&sub_category_id=0&page=1"
current_page = 1

while current_page < 200:
    print(current_page)
    url = base_url + str(current_page)
    #current_page += 1
    r = requests.get(url)
    zute_soup = BeautifulSoup(r.text, 'html.parser')
    firme = zute_soup.findAll('div', {'class': 'jobs-item'})

    for title in firme:
        title1 = title.findAll('h6')[0].text
        print(title1)
        adresa = title.findAll('div', {'class': 'description'})[0].text
        print(adresa)
        kontakt = title.findAll('div', {'class': 'description'})[1].text
        print(kontakt)
        print('\n')
        page_line = "{title1}\n{adresa}\n{kontakt}".format(
            title1=title1,
            adresa=adresa,
            kontakt=kontakt
        )
    current_page += 1

Answer 1

获取 CSV 的一种简单方法是打印以逗号分隔的每一行，然后使用操作系统的“>”写入文件。

import csv
import requests
from bs4 import BeautifulSoup

base_url = "http://www.privredni-imenik.com/pretraga?abcd=&keyword=&cities_id=0&category_id=0&sub_category_id=0&page=1"
current_page = 1


with open('scrape_results.csv', 'w', newline='') as scrape_results:
    csvwriter = csv.writer(scrape_results)

    while current_page < 200:
        url = base_url + str(current_page)
        r = requests.get(url)
        zute_soup = BeautifulSoup(r.text, 'html.parser')
        firme = zute_soup.findAll('div', {'class': 'jobs-item'})

        for title in firme:
            title1 = title.findAll('h6')[0].text
            adresa = title.findAll('div', {'class': 'description'})[0].text
            kontakt = title.findAll('div', {'class': 'description'})[1].text
            csvwriter.writerow([current_page, title1, adresa, kontakt])

        current_page += 1