3

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。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())

这将依次打开每个 browser_type。如果我想并行执行，我将如何处理？如果我想对 url 列表做类似的事情，我会怎么做？

我试过这样做：

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()

但这给了我以下错误：

92: RuntimeWarning: coroutine 'go_to_url' was never awaited
  go_to_url()
RuntimeWarning: Enable tracemalloc to get the object allocation traceback
4

1 回答 1

1

我相信您需要go_to_url使用相同的配方调用您的函数：

asyncio.get_event_loop().run_until_complete(go_to_url())
于 2020-11-03T14:24:07.157 回答