Django print in template all arugment passed from view -

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I would like to quickly debug the purpose that to be able to print all the logic passed from the scene, there are some magic templatetag Is it possible that my time can be saved for writing my job clearly in the curly bracket for all the names?

UPDATE:

I found {% debug%} but print is not necessary that it can be filtered in some way?

Did you see Django Debug Toolbar?

You can set it up for development and see template references which will tell you what you want.

There are also many other useful tools.



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