我想使用 Playwright for Python 一次打开多个 url。但我正在努力弄清楚如何做。这是来自异步文档:
async def main():
async with async_playwright() as p:
for browser_type in [p.chromium, p.firefox, p.webkit]:
browser = await browser_type.launch()
page = await browser.newPage()
await page.goto("https://scrapingant.com/")
await page.screenshot(path=f"scrapingant-{browser_type.name}.png")
await browser.close()
asyncio.get_event_loop().run_until_complete(main())
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这将按顺序打开每个 browser_type。如果我想并行进行,我该怎么做?如果我想对网址列表做类似的事情,我该怎么做?
我尝试这样做:
urls = [
"https://scrapethissite.com/pages/ajax-javascript/#2015",
"https://scrapethissite.com/pages/ajax-javascript/#2014",
]
async def main(url):
async with async_playwright() as p:
browser = await p.chromium.launch(headless=False)
page = await browser.newPage()
await page.goto(url)
await browser.close()
async def go_to_url():
tasks = [main(url) for url in urls]
await asyncio.wait(tasks)
go_to_url()
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但这给了我以下错误:
92: RuntimeWarning: …
Run Code Online (Sandbox Code Playgroud) python webautomation web-scraping playwright playwright-python