I'm trying to create a service to stream some files to clients from the server. However, instead of a URL like this:
$ curl http://localhost:8000/get_file?path=file_name
the client requests the file like this:
$ curl http://localhost:8000/get_file/file_name
Is this possible in Spyne?
Looking at file_soap_http, I wrote the server as such:
# server.py
from spyne import (
Service, rpc, Application,
String,
ByteArray,
)
from spyne.protocol.http import HttpRpc
from spyne.protocol.soap import Soap11
from spyne.server.wsgi import WsgiApplication
from wsgiref.simple_server import make_server
class FileService(Service):
@rpc(String, _returns=ByteArray)
def get_file(ctx, file_name):
print(file_name)
application = Application(
[FileService],
'FileService',
in_protocol=HttpRpc(),
out_protocol=Soap11(),
)
wsgi_application = WsgiApplication(application)
if __name__ == '__main__':
server = make_server('0.0.0.0', 8000, wsgi_application)
server.serve_forever()
This works well with URLs like /get_file?path=file_name, but for URLs like /get_file/file_name, it gives Requested resource not found error:
<soap11env:Envelope xmlns:soap11env="http://schemas.xmlsoap.org/soap/envelope/">
<soap11env:Body>
<soap11env:Fault>
<faultcode>soap11env:Client.ResourceNotFound</faultcode>
<faultstring>Requested resource '{FileService}file_name' not found</faultstring>
<faultactor></faultactor>
</soap11env:Fault>
</soap11env:Body>
</soap11env:Envelope>
I cannot change the fact that the client requests like this. How can I achieve this?
This is doable with HttpPattern.
First have a look at this:
Once you run the daemon, go to:
Or:
etc.
In similar vein, here's how your code should look:
class FileService(Service):
@rpc(String, _returns=ByteArray,
_patterns=[HttpPattern('/get_file/<file_name>')])
def get_file(ctx, file_name):
print(file_name)