Reputation: 712
I am a PHP developer and I used to get query string using $_SERVER['QUERY_STRING']
in PHP.
What is the Python 2.7 syntax for this?
import web
import speech_recognition as sr
from os import path
urls = (
'/voice', 'Voice'
)
app = web.application(urls, globals())
class Voice:
def GET(self):
WAV_FILE = path.join(path.dirname(path.realpath("C:\Python27")),'wavfile.wav')
r = sr.Recognizer()
with sr.WavFile("C:\Python27\wavfile.wav") as source:
audio = r.record(source) # read the entire WAV file
output = r.recognize_google(audio)
return output
if __name__ == "__main__":
app.run()
Upvotes: 0
Views: 375
Reputation: 703
http://webpy.org/cookbook/input
user_data = web.input()
Or use the urlparse library:
https://docs.python.org/2/library/urlparse.html
from urlparse import urlparse
o = urlparse('http://www.cwi.nl:80/%7Eguido?x=y')
Upvotes: 1
Reputation: 136995
Assuming you are using web.py
(which your code suggests) you can use web.ctx.query
(which includes the ?
) or web.ctx.env['QUERY_STRING']
, which doesn't:
import web
urls = (
'/', 'index',
)
class index:
def GET(self):
return "web.ctx.env['QUERY_STRING']: {}".format(
web.ctx.env['QUERY_STRING'])
if __name__ == '__main__':
app = web.application(urls, globals())
app.run()
See the cookbook entry on ctx
for more.
Upvotes: 0
Reputation: 178
import urlparse
url = 'http://example.com/?q=abc&p=123'
par = urlparse.parse_qs(urlparse.urlparse(url).query)
Upvotes: 0