Opened 17 years ago
Closed 15 years ago
#8928 closed New feature (duplicate)
Make a WSGI compatibility layer for Django middleware
| Reported by: | simon | Owned by: | nobody |
|---|---|---|---|
| Component: | HTTP handling | Version: | 1.0 |
| Severity: | Normal | Keywords: | |
| Cc: | andy@…, dane.springmeyer@… | Triage Stage: | Accepted |
| Has patch: | no | Needs documentation: | no |
| Needs tests: | no | Patch needs improvement: | no |
| Easy pickings: | no | UI/UX: | no |
Description
Another Mark Ramm suggestion at DjangoCon.
Change History (9)
comment:1 by , 17 years ago
comment:2 by , 17 years ago
Ok, previous comment was meant for ticket 8927.
In this case, it would be awesome if the distinction between django middleware and wsgi middleware could mostly go away.
comment:3 by , 17 years ago
| Cc: | added |
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comment:5 by , 17 years ago
| Triage Stage: | Unreviewed → Accepted |
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comment:6 by , 17 years ago
| Cc: | added |
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comment:7 by , 17 years ago
| Component: | Uncategorized → HTTP handling |
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comment:8 by , 15 years ago
| Severity: | → Normal |
|---|---|
| Type: | → New feature |
comment:9 by , 15 years ago
| Easy pickings: | unset |
|---|---|
| Resolution: | → duplicate |
| Status: | new → closed |
This appears to be a dupe of #8927.
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The idea here is that there should be easy access to a wsgi environ like dictionary, to make calling wsgi applications from within django trivial.
I would even propose that making a call_wsgi_app function that grabs the environ, creates a start_response callable, calls the wsgi app, and returns the response properly.
This would let users write code like:
def someview(request, *args) do_something(args) return call_wsgi_app(wsgi_app, request)This would allow people who use django with SQLAlchemy to user RUM (A django admin like wsgi app) from a veiw, or to use some SOAP wsgi application, or a TurboGears app, or whatever they want in a Django view.