I have a bunch of files in wav. I made a simple script to convert them to flac so I can use it with the google speech api. Here is the python code:
import urllib2
url = "https://www.google.com/speech-api/v1/recognize?client=chromium&lang=en-US"
audio = open('somefile.flac','rb').read()
headers={'Content-Type': 'audio/x-flac; rate=16000', 'User-Agent':'Mozilla/5.0'}
request = urllib2.Request(url, data=audio, headers=headers)
response = urllib2.urlopen(request)
print response.read()
However I am getting this error:
Traceback (most recent call last):
File "transcribe.py", line 7, in <module>
response = urllib2.urlopen(request)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 392, in open
response = self._open(req, data)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 410, in _open
'_open', req)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 370, in _call_chain
result = func(*args)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1194, in https_open
return self.do_open(httplib.HTTPSConnection, req)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1161, in do_open
raise URLError(err)
urllib2.URLError: <urlopen error [Errno 32] Broken pipe>
I thought at first that it was because the file is too big. But I recorded myself for 5 seconds and it still does the same.
I dont think google ha released the api yet so it's hard to understand why its failing.
Is there any other good speech-to-text api out there that can be used in either Python or Node?
----- Editing for my attempt with requests:
import json
import requests
url = 'https://www.google.com/speech-api/v1/recognize?client=chromium&lang=en-US'
data = {'file': open('file.flac', 'rb')}
headers = {'Content-Type': 'audio/x-flac; rate=16000', 'User-Agent':'Mozilla/5.0'}
r = requests.post(url, data=data, headers=headers)
# r = requests.post(url, files=data, headers=headers) ## does not work either
# r = requests.post(url, data=open('file.flac', 'rb').read(), headers=headers) ## does not work either
print r.text
Produced the same problem as above.
The API Accepts HTTP POST requests. You're using a HTTP GET Request here. This can be confirmed by loading the URI in your code directly into a browser:
HTTP method GET is not supported by this URL
Error 405
Also, i'd recommend using the requests python library. See http://www.python-requests.org/en/latest/user/quickstart/#post-a-multipart-encoded-file
Lastly, it seems that the API only accepts segments up to 15 seconds long. Perhaps your error is the file is too large? If you can upload an example flac file, perhaps we could diagnose further.
Related
I have been searching for a way to implement 2 legged oauth in python 3 to work with the brightcloud api. They offer several code examples using java, php, ruby, .NET C# here: https://bcws.brightcloud.com/code-samples.php. I tried following the same logic to convert the java example into python, however, I'm relatively new to python and I quickly came unstuck.
I have tried implementing using rauth, however, the basic setup utilises a request_token_url which is not provided by brightcloud. I also tried implementing with the following code which was based on this answer -
How do I send a POST using 2-legged oauth2 in python?
import oauth2
import urllib #for url-encode
import urllib.request
import time #Unix timestamp import oauth2
# construct request url
base_url = "http://thor.brightcloud.com/rest"
uri_info_path = "/uris"
url = urllib.parse.quote_plus("http://www.booking.com")
# api key and secret
consumer_key = 'MY_CONSUMER_KEY'
consumer_secret = 'MY_CONSUMER_SECRET'
# contruct endpoint
endpoint = rest_endpoint + uri_info_path + '/' + url
# build request
def build_request(url, method):
params = {
'oauth_version': "1.0",
'oauth_nonce': oauth2.generate_nonce(),
'oauth_timestamp': int(time.time())
}
consumer = oauth2.Consumer(key=consumer_key, secret=consumer_secret)
params['oauth_consumer_key'] = consumer.key
req = oauth2.Request(method=method, url=url, parameters=params)
signature_method = oauth2.SignatureMethod_HMAC_SHA1()
req.sign_request(signature_method, consumer, None)
return req
# call
request = build_request(endpoint,'GET')
u = urllib.request.urlopen(request.to_url())
data = u.read()
print (data)
There is a problem with this line:
u = urllib.request.urlopen(request.to_url())
Which generates the following traceback:
Traceback (most recent call last):
File "bright.py", line 37, in
u = urllib.request.urlopen(request.to_url())
File "/usr/lib/python3.5/urllib/request.py", line 163, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.5/urllib/request.py", line 472, in open
response = meth(req, response)
File "/usr/lib/python3.5/urllib/request.py", line 582, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.5/urllib/request.py", line 510, in error
return self._call_chain(*args)
File "/usr/lib/python3.5/urllib/request.py", line 444, in _call_chain
result = func(*args)
File "/usr/lib/python3.5/urllib/request.py", line 590, in >http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 401: Unauthorized
Any help would be much appreciated.
I learned how to download a picture from a certain URL with python as:
import urllib
imgurl="http://www.digimouth.com/news/media/2011/09/google-logo.jpg"
resource = urllib.urlopen(imgurl)
output = open("test.jpg","wb")
output.write(resource.read())
output.close()
and it worked well, but when i changed the URL to
imgurl="http://farm1.static.flickr.com/96/242125326_607a826afe_o.jpg"
it did not work, and gave the information
File "face_down.py", line 3, in <module>
resource = urllib2.urlopen(imgurl)
File "D:\Python27\another\Lib\urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "D:\Python27\another\Lib\urllib2.py", line 431, in open
response = self._open(req, data)
File "D:\Python27\another\Lib\urllib2.py", line 449, in _open
'_open', req)
File "D:\Python27\another\Lib\urllib2.py", line 409, in _call_chain
result = func(*args)
File "D:\Python27\another\Lib\urllib2.py", line 1227, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "D:\Python27\another\Lib\urllib2.py", line 1197, in do_open
raise URLError(err)
urllib2.URLError: <urlopen error [Errno 10060] >
and I tried to open the latter image URL, and it could be shown as the former, I have no idea to solve it~~ help~~~~
You can try using requests module. The response will be some bytes. So, you can iterate over those byte chunks and write to the file.
import requests
url = "http://farm1.static.flickr.com/96/242125326_607a826afe_o.jpg"
r = requests.get(url)
path = "filename.jpg"
with open(path, 'wb') as f:
for chunk in r:
f.write(chunk)
I looked up both of the addresses and the second one does not lead anywhere. That is probably the problem.
import urllib
imgurl="webpage url"
openimg = urllib.urlopen(imgurl) #opens image (prepares it)
img = open("test.jpg","wb") #opens the img to read it
img.write(openimg.read()) #prints it to the console
img.close() #closes img
Try the link again in your webpage and if it turns up with "webpage not available" that is probably the problem.
I was trying to download images with url's that change but got an error.
url_image="http://www.joblo.com/timthumb.php?src=/posters/images/full/"+str(title_2)+"-poster1.jpg&h=333&w=225"
user_agent = 'Mozilla/5.0 (Windows NT 6.1; Win64; x64)'
headers = {'User-Agent': user_agent}
req = urllib.request.Request(url_image, None, headers)
print(url_image)
#image, h = urllib.request.urlretrieve(url_image)
with urllib.request.urlopen(req) as response:
the_page = response.read()
#print (the_page)
with open('poster.jpg', 'wb') as f:
f.write(the_page)
Traceback (most recent call last):
File "C:\Users\luke\Desktop\scraper\imager finder.py", line 97, in
with urllib.request.urlopen(req) as response:
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 162, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 465, in open
response = self._open(req, data)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 483, in _open
'_open', req)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 443, in _call_chain
result = func(*args)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1268, in http_open
return self.do_open(http.client.HTTPConnection, req)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1243, in do_open
r = h.getresponse()
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1174, in getresponse
response.begin()
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 282, in begin
version, status, reason = self._read_status()
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 264, in _read_status
raise BadStatusLine(line)
http.client.BadStatusLine:
My advice is to use urlib2. In addition, I've written a nice function (I think) that will also allow gzip encoding (reduce bandwidth) if the server supports it. I use this for downloading social media files, but should work for anything.
I would try to debug your code, but since it's just a snippet (and the error messages are formatted badly), it's hard to know exactly where your error is occurring (it's certainly not line 97 in your code snippet).
This isn't as short as it could be, but it's clear and reusable. This is python 2.7, it looks like you're using 3 - in which case you google some other questions that address how to use urllib2 in python 3.
import urllib2
import gzip
from StringIO import StringIO
def download(url):
"""
Download and return the file specified in the URL; attempt to use
gzip encoding if possible.
"""
request = urllib2.Request(url)
request.add_header('Accept-Encoding', 'gzip')
try:
response = urllib2.urlopen(request)
except Exception, e:
raise IOError("%s(%s) %s" % (_ERRORS[1], url, e))
payload = response.read()
if response.info().get('Content-Encoding') == 'gzip':
buf = StringIO(payload)
f = gzip.GzipFile(fileobj=buf)
payload = f.read()
return payload
def save_media(filename, media):
file_handle = open(filename, "wb")
file_handle.write(media)
file_handle.close()
title_2 = "10-cloverfield-lane"
media = download("http://www.joblo.com/timthumb.php?src=/posters/images/full/{}-poster1.jpg&h=333&w=225".format(title_2))
save_media("poster.jpg", media)
I try to write a simple python3 script that gets some playlist informations via the youtube API. However I always get a 401 Error whereas it works perfectly when I enter the request string in a browser or making a request with w-get. I'm relatively new to python and I guess I'm missing some important point here.
This is my script. Of course I actually use a real API-Key.
from urllib.request import Request, urlopen
from urllib.parse import urlencode
api_key = "myApiKey"
playlist_id = input('Enter playlist id: ')
output_file = input('Enter name of output file (default is playlist id')
if output_file == '':
output_file = playlist_id
url = 'https://www.googleapis.com/youtube/v3/playlistItems'
params = {'part': 'snippet',
'playlistId': playlist_id,
'key': api_key,
'fields': 'items/snippet(title,description,position,resourceId/videoId),nextPageToken,pageInfo/totalResults',
'maxResults': 50,
'pageToken': '', }
data = urlencode(params)
request = Request(url, data.encode('utf-8'))
response = urlopen(request)
content = response.read()
print(content)
Unfortunately it rises a error at response = urlopen(request)
Traceback (most recent call last):
File "gpd-helper.py", line 35, in <module>
response = urlopen(request)
File "/usr/lib/python3.4/urllib/request.py", line 153, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.4/urllib/request.py", line 461, in open
response = meth(req, response)
File "/usr/lib/python3.4/urllib/request.py", line 571, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.4/urllib/request.py", line 499, in error
return self._call_chain(*args)
File "/usr/lib/python3.4/urllib/request.py", line 433, in _call_chain
result = func(*args)
File "/usr/lib/python3.4/urllib/request.py", line 579, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 401: Unauthorized
I looked up the documentation but couldn't find any hint. According to the docs other authentication than the api key is not required for listing a public playlist.
After diving deeper into the docs of python and google I found the solution to my problem.
Pythons Request object automatically creates a POST request when the data parameter is given but the youtube api expects GET (with post params)
The Solution is to ether supply the GET argument for the method parameter in python 3.4
request = Request(url, data.encode('utf-8'), method='GET')
or concatenate the url with the urlencoded post data
request = Request(url + '?' + data)
I have the following python script and it works beautifully.
import urllib2
url = 'http://abc.com' # write the url here
usock = urllib2.urlopen(url)
data = usock.read()
usock.close()
print data
however, some of the URL's I give it may redirect it 2 or more times. How can I have python wait for redirects to complete before loading the data.
For instance when using the above code with
http://www.google.com/search?hl=en&q=KEYWORD&btnI=1
which is the equvilant of hitting the im lucky button on a google search, I get:
>>> url = 'http://www.google.com/search?hl=en&q=KEYWORD&btnI=1'
>>> usick = urllib2.urlopen(url)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
response = meth(req, response)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
'http', request, response, code, msg, hdrs)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 438, in error
return self._call_chain(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
result = func(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 521, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden
>>>
Ive tried the (url, data, timeout) however, I am unsure what to put there.
EDIT:
I actually found out if I dont redirect and just used the header of the first link, I can grab the location of the next redirect and use that as my final link
Use requests as the other answer states, here is an example. The redirect will be in r.url. In the example below the http is redirected to https
For HEAD:
In [1]: import requests
...: r = requests.head('http://github.com', allow_redirects=True)
...: r.url
Out[1]: 'https://github.com/'
For GET:
In [1]: import requests
...: r = requests.get('http://github.com')
...: r.url
Out[1]: 'https://github.com/'
Note for HEAD you have to specify allow_redirects, if you don't you can get it in the headers but this is not advised.
In [1]: import requests
In [2]: r = requests.head('http://github.com')
In [3]: r.headers.get('location')
Out[3]: 'https://github.com/'
To download the page you will need GET, you can then access the page using r.content
You might be better off with Requests library which has better APIs for controlling redirect handling:
https://requests.readthedocs.io/en/master/user/quickstart/#redirection-and-history
Requests:
https://pypi.org/project/requests/ (urllib replacement for humans)