I am using recombees API for a recommendation and there is a batch method to send all the user data to the API.
The code for the following:
for i in range(0,len(list_of_ratings)):
name = str(list_of_ratings[i].user)
series = str(list_of_ratings[i].series)
rate = list_of_ratings[i].rating
print(name + ' ' + series + ' ' + str(rate))
request = AddRating(name, series, rate ,cascade_create=True)
requests.append(request)
try:
client.send(Batch(requests))
except APIException as e:
print(e)
except ResponseException as e:
print(e)
except ApiTimeoutException as e:
print(e)
except Exception as e:
print(e)
But the problem is it does not send all the data. There are 946 data objects that I have in a Django model but the first time when i ran this only 20 were sent and during the 2nd time only 6.
I dont know whats causing the issue.
Any help is appreciated.
Perhaps there is some error in your batch. I would suggest printing the batch result to see eventual error messages:
res = client.send(Batch(requests))
print(res)
I'm trying to write a script for barcode printers.
Here is my Code:
def printLabel(barcode, qty, articlenr=''):
dev = findPrinter()
dev.write("")
return True
if __name__ =="__main__":
for i in range(2):
try:
# Everything works fine, the for loop breaks
printLabel('01234567890123','1', 'Hello World')
break
except usb.core.USBError as error:
if "Resource busy" in unicode(error):
# Printer is doing something, wait 3 seconds, try again
time.sleep(3)
print 'Retrying after error:"Resource busy"\n' + str(error)
printLabel('01234567890123','1', 'Hello World')
elif "Operation Timeout" in unicode(error):
# Printer is not responding
print 'ERROR! Check Paper!\n' + str(error)
else:
print 'Unknown error' + str(error)
I have to ask for the status of the printer before he tries to printLabel, because if I can't ask for it he maybe gets in the "Resource busy" if clause inside the exception, prints the label, starts all over again and prints it again.
But if I could ask for the status I can say him something like:
def waitForPrinter(dev):
dev = findPrinter()
if dev.status() is True: ## True if he is ready and False if he is Busy or something
printLabel(arguments and stuff)
else:
time.sleep(3)
waitForPrinter()
I couldn't find a function in the documentation for asking the device status and don't know how to solve the problem right now.
At the moment i can only fetch the device status with the USBError.
I wrote a script to upload a video to YouTube using YouTube Data API v3 in the python with help of example given in Example code.
And I wrote another script to add uploaded video to playlist using same YouTube Data API v3 you can be seen here
After that I wrote a single script to upload video and add that video to playlist. In that I took care of authentication and scops still I am getting permission error. here is my new script
#!/usr/bin/python
import httplib
import httplib2
import os
import random
import sys
import time
from apiclient.discovery import build
from apiclient.errors import HttpError
from apiclient.http import MediaFileUpload
from oauth2client.file import Storage
from oauth2client.client import flow_from_clientsecrets
from oauth2client.tools import run
# Explicitly tell the underlying HTTP transport library not to retry, since
# we are handling retry logic ourselves.
httplib2.RETRIES = 1
# Maximum number of times to retry before giving up.
MAX_RETRIES = 10
# Always retry when these exceptions are raised.
RETRIABLE_EXCEPTIONS = (httplib2.HttpLib2Error, IOError, httplib.NotConnected,
httplib.IncompleteRead, httplib.ImproperConnectionState,
httplib.CannotSendRequest, httplib.CannotSendHeader,
httplib.ResponseNotReady, httplib.BadStatusLine)
# Always retry when an apiclient.errors.HttpError with one of these status
# codes is raised.
RETRIABLE_STATUS_CODES = [500, 502, 503, 504]
CLIENT_SECRETS_FILE = "client_secrets.json"
# A limited OAuth 2 access scope that allows for uploading files, but not other
# types of account access.
YOUTUBE_UPLOAD_SCOPE = "https://www.googleapis.com/auth/youtube.upload"
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
# Helpful message to display if the CLIENT_SECRETS_FILE is missing.
MISSING_CLIENT_SECRETS_MESSAGE = """
WARNING: Please configure OAuth 2.0
To make this sample run you will need to populate the client_secrets.json file
found at:
%s
with information from the APIs Console
https://code.google.com/apis/console#access
For more information about the client_secrets.json file format, please visit:
https://developers.google.com/api-client-library/python/guide/aaa_client_secrets
""" % os.path.abspath(os.path.join(os.path.dirname(__file__),
CLIENT_SECRETS_FILE))
def get_authenticated_service():
flow = flow_from_clientsecrets(CLIENT_SECRETS_FILE, scope=YOUTUBE_UPLOAD_SCOPE,
message=MISSING_CLIENT_SECRETS_MESSAGE)
storage = Storage("%s-oauth2.json" % sys.argv[0])
credentials = storage.get()
if credentials is None or credentials.invalid:
credentials = run(flow, storage)
return build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
http=credentials.authorize(httplib2.Http()))
def initialize_upload(title,description,keywords,privacyStatus,file):
youtube = get_authenticated_service()
tags = None
if keywords:
tags = keywords.split(",")
insert_request = youtube.videos().insert(
part="snippet,status",
body=dict(
snippet=dict(
title=title,
description=description,
tags=tags,
categoryId='26'
),
status=dict(
privacyStatus=privacyStatus
)
),
# chunksize=-1 means that the entire file will be uploaded in a single
# HTTP request. (If the upload fails, it will still be retried where it
# left off.) This is usually a best practice, but if you're using Python
# older than 2.6 or if you're running on App Engine, you should set the
# chunksize to something like 1024 * 1024 (1 megabyte).
media_body=MediaFileUpload(file, chunksize=-1, resumable=True)
)
vid=resumable_upload(insert_request)
#Here I added lines to add video to playlist
#add_video_to_playlist(youtube,vid,"PL2JW1S4IMwYubm06iDKfDsmWVB-J8funQ")
#youtube = get_authenticated_service()
add_video_request=youtube.playlistItems().insert(
part="snippet",
body={
'snippet': {
'playlistId': "PL2JW1S4IMwYubm06iDKfDsmWVB-J8funQ",
'resourceId': {
'kind': 'youtube#video',
'videoId': vid
}
#'position': 0
}
}
).execute()
def resumable_upload(insert_request):
response = None
error = None
retry = 0
vid=None
while response is None:
try:
print "Uploading file..."
status, response = insert_request.next_chunk()
if 'id' in response:
print "'%s' (video id: %s) was successfully uploaded." % (
title, response['id'])
vid=response['id']
else:
exit("The upload failed with an unexpected response: %s" % response)
except HttpError, e:
if e.resp.status in RETRIABLE_STATUS_CODES:
error = "A retriable HTTP error %d occurred:\n%s" % (e.resp.status,
e.content)
else:
raise
except RETRIABLE_EXCEPTIONS, e:
error = "A retriable error occurred: %s" % e
if error is not None:
print error
retry += 1
if retry > MAX_RETRIES:
exit("No longer attempting to retry.")
max_sleep = 2 ** retry
sleep_seconds = random.random() * max_sleep
print "Sleeping %f seconds and then retrying..." % sleep_seconds
time.sleep(sleep_seconds)
return vid
if __name__ == '__main__':
title="sample title"
description="sample description"
keywords="keyword1,keyword2,keyword3"
privacyStatus="public"
file="myfile.mp4"
vid=initialize_upload(title,description,keywords,privacyStatus,file)
print 'video ID is :',vid
I am not able to figure out what is wrong. I am getting permission error. both script works fine independently.
could anyone help me figure out where I am wrong or how to achieve uploading video and adding that too playlist.
I got the answer actually in both the independent script scope is different.
scope for uploading is "https://www.googleapis.com/auth/youtube.upload"
scope for adding to playlist is "https://www.googleapis.com/auth/youtube"
as scope is different so I had to handle authentication separately.
I have a GUI created using PyQt. In the GUI their is a button which when pressed send some data to client. Following is my code
class Main(QtGui.QTabWidget, Ui_TabWidget):
def __init__(self):
QtGui.QTabWidget.__init__(self)
self.setupUi(self)
self.pushButton_8.clicked.connect(self.updateActual)
def updateActual():
self.label_34.setText(self.comboBox_4.currentText())
HOST = '127.0.0.1' # The remote host
PORT = 8000 # The same port as used by the server
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
try:
s.connect((displayBoard[str(self.comboBox_4.currentText())], PORT))
except socket.error as e:
err1 = str(self.comboBox_4.currentText()) + " is OFF-LINE"
reply2 = QtGui.QMessageBox.critical(self, 'Error', err1, QtGui.QMessageBox.Ok)
if reply2 == QtGui.QMessageBox.Ok:
pass #stop execution at this point
fileName = str(self.comboBox_4.currentText()) + '.txt'
f = open(fileName)
readLines = f.readlines()
line1 = int(readLines[0])
f.close()
Currently if a user clicks 'ok' in QMessageBox the program will continue code execution in case their is socket exception. Thus my question is how can I stop the execution of code after 'except' in a clean way such that my UI doesn't crash and user can continue using it?
Yes, you can simply return from the if block:
if reply2 == QtGui.QMessageBox.Ok:
return
Alternatively, move your code for when it doesn't raise socket.error into an else block:
try: # this might fail
s.connect(...)
except socket.error as e: # what to do if it fails
err1 = ...
reply2 = QtGui.QMessageBox.critical(...)
else: # what to do if it doesn't
with open(fileName) as f:
line1 = int(f.readline().strip())
Note that:
You don't actually need to deal with the return from the message box, as it could only be OK and you have no else option;
you should generally use with for file handling, it will automatically close at the end of the block; and
you can simplify your file handling code by only reading the first line.
Ok, I've tried to resolve this with a couple of different libraries. I'm working on a script to look at thousands of sites and kick out specific items on the pages. I need to be able to reset the connection so that the script will continue without losing any data. I've tried catching the error and waiting but that doesn't seem to fix it as it eventually causes the script to error completely out. I get the error on the below snippet of code in my status check module.
def status(url): #checks the response code
try:
req=urllib2.urlopen(url)
response=req.getcode()
return response
except urllib2.HTTPError, e:
return e.code
print e.code
except urllib2.URLError, e:
print e.args
return e.args
But before trying this I used the below as instead of urrlib2
parsedurl = urlparse(url)
conn = httplib.HTTPConnection(parsedurl.netloc)
conn.request('HEAD',parsedurl.path)
response = conn.getresponse()
return response.status