转到下一页，但它不会使用Selenium和Scrapy刮取元素

Question

我正在尝试使用Selenium刮取所有页面并单击“下一页”按钮。然而，当我移动到下一页时，URL并没有改变。我可以移动到所有页面，但我只得到从第一页刮下来的项目，不知道如何使其适用于所有页面。 对我应该做什么有什么建议吗

提前谢谢你

守则：

class MilieuProperties(scrapy.Spider):
    name = 'milieu_properties'
    start_urls = [
        # FOR SALE
        'https://www.milieuproperties.com/search-results.aspx?paramb=ADVANCE%20SEARCH:%20Province%20(Western%20Cape),%20%20Area%20(Cape%20Town)',
        'https://www.milieuproperties.com/RentalByCategory.aspx'
    ]

    def __init__(self):


        
        #headless options
        options = Options()
        options.add_argument('--no-sandbox')
        options.add_argument("--headless")
        options.add_experimental_option("excludeSwitches", ["enable-automation"])
        options.add_experimental_option('useAutomationExtension', False)
        self.driver = webdriver.Chrome('path',options=options)

    
    def parse(self,response):
        self.driver.get(response.url)
        current_page_number = self.driver.find_element_by_css_selector('#ContentPlaceHolder1_lvDataPager1>span').text
        while True:
            try: 
                elem = WebDriverWait(self.driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//*[@id="ContentPlaceHolder1_lvDataPager1"]/a[text()="Next" and not(@class)]')))
                elem.click()
            except TimeoutException:
                break
            WebDriverWait(self.driver, 10).until(lambda driver: self.driver.find_element_by_css_selector('#ContentPlaceHolder1_lvDataPager1>span').text != current_page_number)
            current_page_number = self.driver.find_element_by_css_selector('#ContentPlaceHolder1_lvDataPager1>span').text


        offering = response.css('span#ContentPlaceHolder1_lblbreadcum::text').get()
        try:
            offering = 'rent' if 'Rental' in offering else 'buy'
        except TypeError:
            offering = 'buy'

        base_link = response.request.url.split('/')
        try:
            base_link = base_link[0] + '//' + base_link[2] + '/'
        except:
            pass

        for p in response.xpath('//div[@class="ct-itemProducts ct-u-marginBottom30 ct-hover"]'):
            link = base_link + p.css('a::attr(href)').get()

            yield scrapy.Request(
                link,
                callback=self.parse_property,
                meta={'item': {
                    'url': link,
                    'offering': offering,
                    }},
            )


    def parse_property(self, response):
        item = response.meta.get('item')
        . . .

Answer 1

您可以不使用Scrapy获取数据。请尝试以下代码：

from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC

driver = webdriver.Chrome()

links = []
url = 'https://www.milieuproperties.com/search-results.aspx?paramb=ADVANCE%20SEARCH:%20Province%20(Western%20Cape),%20%20Area%20(Cape%20Town)'
driver.get(url)
current_page_number = driver.find_element_by_css_selector('#ContentPlaceHolder1_lvDataPager1>span').text
while True:
    links.extend([link.get_attribute('href') for link in driver.find_elements_by_css_selector('.hoverdetail a')])
    try: 
        elem = WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//*[@id="ContentPlaceHolder1_lvDataPager1"]/a[text()="Next" and not(@class)]')))
        elem.click()
    except TimeoutException:
        break
    WebDriverWait(driver, 10).until(lambda driver: driver.find_element_by_css_selector('#ContentPlaceHolder1_lvDataPager1>span').text != current_page_number)
    current_page_number = driver.find_element_by_css_selector('#ContentPlaceHolder1_lvDataPager1>span').text

print(links)