CODEWITHSUNDEEP
pythonregexurltornado
ManuParra
ManuParra

Reputation: 1531

Get full url from tornado web patterns

I use python Tornado, and I want extract from this URL:

http://myserver.com/v1/files_get/app/path/for/you/myfile.txt

this value:

app

and the rest of path (path): 

/path/for/you/myfile.txt

I use this REGEXP for it into Tornado:

tornado.web.Application([
    (r"/v1/files_get/(?P<root>[^\/]+)/?(?P<path>[^\/]+)?/?", files_get),
])

I don't know if it is possible, because my regexp only get only just level of path.

¿What should be the correct regex for it?

Upvotes: 0

Views: 552

Answers (1)

traybold
traybold

Reputation: 444

Try this:

r'/v1/files_get/(?P<root>[^/]+)(?P<path>/.*)?'

/ is not magic, and you're not using it as a delimiter (as you would in Perl or Ruby), so it does not need to be escaped.

Since you want to capture everything after <root>, if it exists, you can make that whole group optional.

Upvotes: 1

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