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在 FLASK 中运行 pypupeteer 会产生 ValueError:信号仅适用于主线程

我正在尝试将 pyppeteer 集成到烧瓶应用程序中。我有运行 pyppeteer 并截取页面截图的 python 脚本。如果我单独运行脚本,这是工作文件。

问题是相同的脚本时,我在烧瓶APP运行它不起作用。

我收到以下错误:

loop.run_until_complete(capture(url, 123123))
File "/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/
Versions/3.7/lib/python3.7/asyncio/base_events.py", 
line 568, in run_until_complete
return future.result()
File "/App-path/flaskr/image_capture/__init__.py", line 6, in capture
browser = await launch()
File "/usr/local/lib/python3.7/site-packages/pyppeteer/launcher.py", 
line 311, in launch
return await Launcher(options, **kwargs).launch()
File "/usr/local/lib/python3.7/site-packages/pyppeteer/launcher.py", 
line 180, in launch
signal.signal(signal.SIGINT, _close_process)
File"/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/
Versions/3.7/lib/python3.7/signal.py", line 47, in signal
handler = _signal.signal(_enum_to_int(signalnum), 
_enum_to_int(handler))
ValueError: signal only works in main thread.
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以下代码用于捕获屏幕截图。

async def capture(code_url, codeId):
browser = await launch()
# print('Hello')
page = …
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flask python-3.x google-chrome-headless puppeteer pyppeteer

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1
解决办法
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