使用scrapy-playwright和only playwright的网站工作方式不同

kkbh8khc  于 2022-11-23  发布在  其他
关注(0)|答案(1)|浏览(212)

我尝试使用scrapy-playwright登录一个网页,因为我想与scrapy很好地集成.我不能使用scrapy-playwright登录,因为它会重定向到一个不存在的页面.我也尝试过做一个帖子请求,而不是点击,这也不工作.
然而,如果我尝试同样的事情只使用剧作家,它的工作完美...有没有不同的网站打开与scrapy-playwright相比,只有剧作家?有谁知道如何解决这个问题使用scrapy-playwright?
杂剧作家守则:

def start_requests(self):
    yield scrapy.Request(
        url = self.url,
        meta = dict(
            playwright = True,
            playwright_include_page = True,
            playwright_page_methods = [PageMethod('wait_for_selector', 'a[data-toggle=dropdown]')],
                ),
        callback = self.sign_in,
        )

async def sign_in(self, response):
    page = response.meta['playwright_page']
    while await page.is_visible("button[class='close close-news']"):
        await page.click("button[class='close close-news']")
    await page.click('button#declineAllConsentSummary')
    await page.click('div.my-account-sub > a[data-toggle=dropdown]', timeout=10000)
    await page.fill('input#j_username_header', os.getenv(self.usernameKey), timeout=10000)
    await page.fill('input#j_password_header', os.getenv(self.passwordKey), timeout=10000)
    await page.click('button#responsiveMyAccLoginGA')

剧作家代号:

async def test_async_playwright(self):
    async with async_playwright() as playwright:
        browser = await playwright.chromium.launch(headless=False)
        context = await browser.new_context(base_url=self.url)
        page = await context.new_page()
        
        await page.goto(self.url, wait_until='commit')
        while await page.is_visible("button[class='close close-news']"):
            await page.click("button[class='close close-news']")
        await page.click('button#declineAllConsentSummary')
        await page.wait_for_selector('a[data-toggle=dropdown]')
        await page.click('div.my-account-sub > a[data-toggle=dropdown]', timeout=5000)
        await page.fill('input#j_username_header', os.getenv(self.usernameKey), timeout=5000)
        await page.fill('input#j_password_header', os.getenv(self.passwordKey), timeout=5000)
        await page.click('button#responsiveMyAccLoginGA')
kqlmhetl

kqlmhetl1#

作为一种可能的解决方法,如果在令牌/cookie被授予后您被重定向(到损坏的页面),您也可以导航到正常的站点url,并且您应该登录

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