如何在python中获取URL的控制台输出？

Question

I searched for a solution to my problem but couldnt find any.

Is there a way to open a URL in python and get the console output from that url?

Im trying to check a list of URLs for Javascript errors. The only solution I found was using Selenium but its incredible slow.

Answer 1

• For chrome errors you might find the node module of LightHouse suiting your needs.
• For Firefox there is Browser_Console .

From first look the Chrome solution seems easier.

• Last but not least I stumbled over the python module selenium which can do liveserver testing.

For reference see this django use case , there is also an other post how to catch js errors