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  • PyDev and Django: PyDev breaking Django shell?

    - by Rosarch
    I've set up a new project, and populated it with simple models. (Essentially I'm following the tut.) When I run python manage.py shell on the command line, it works fine: >python manage.py shell Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. (InteractiveConsole) >>> from mysite.myapp.models import School >>> School.objects.all() [] Works great. Then, I try to do the same thing in Eclipse (using a Django project that is composed of the same files.) Right click on mysite project Django Shell with Django environment This is the output from the PyDev Console: >>> import sys; print('%s %s' % (sys.executable or sys.platform, sys.version)) C:\Python26\python.exe 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] >>> >>> from django.core import management;import mysite.settings as settings;management.setup_environ(settings) 'path\\to\\mysite' >>> from mysite.myapp.models import School >>> School.objects.all() Traceback (most recent call last): File "<console>", line 1, in <module> File "C:\Python26\lib\site-packages\django\db\models\query.py", line 68, in __repr__ data = list(self[:REPR_OUTPUT_SIZE + 1]) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 83, in __len__ self._result_cache.extend(list(self._iter)) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 238, in iterator for row in self.query.results_iter(): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 287, in results_iter for rows in self.execute_sql(MULTI): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 2368, in execute_sql cursor = self.connection.cursor() File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 81, in cursor cursor = self._cursor() File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 170, in _cursor self.connection = Database.connect(**kwargs) OperationalError: unable to open database file What am I doing wrong here?

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  • force delete row on django app after migration

    - by unsorted
    After a migration with south, I ended up deleting a column. Now the current data in one of my tables is screwed up and I want to delete it, but attempts to delete just result in an error: >>> d = Degree.objects.all() >>> d.delete() Traceback (most recent call last): File "<console>", line 1, in <module> File "C:\Python26\lib\site-packages\django\db\models\query.py", line 440, in d elete for i, obj in izip(xrange(CHUNK_SIZE), del_itr): File "C:\Python26\lib\site-packages\django\db\models\query.py", line 106, in _ result_iter self._fill_cache() File "C:\Python26\lib\site-packages\django\db\models\query.py", line 760, in _ fill_cache self._result_cache.append(self._iter.next()) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 269, in i terator for row in compiler.results_iter(): File "C:\Python26\lib\site-packages\django\db\models\sql\compiler.py", line 67 2, in results_iter for rows in self.execute_sql(MULTI): File "C:\Python26\lib\site-packages\django\db\models\sql\compiler.py", line 72 7, in execute_sql cursor.execute(sql, params) File "C:\Python26\lib\site-packages\django\db\backends\util.py", line 15, in e xecute return self.cursor.execute(sql, params) File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 200, in execute return Database.Cursor.execute(self, query, params) DatabaseError: no such column: students_degree.abbrev >>> Is there a simple way to just force a delete? Do I drop the table and then rerun manage.py schemamigration to recreate the table in south?

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  • how python http request and response works

    - by Apache
    hi expert, i'm newbie for python, i use to learn using sample, i use python to scan wifi to get ssid, and now i want to send the data to the server, then i did as follow import httplib,urllib params = urllib.urlencode({"ssid":"guest"}) headers = {"Content-type":"application/x-www-form-urlencoded","Accept":"text/plain"} conn=httplib.HTTPConnection("http://223.56.124.58:8080/wireless") conn.request("POST","data",params,headers) response = conn.getresponse() print "Response" print response.status print "-----" print response.reason data = response.read() print data conn.close() but when execute the code i'm getting as follow root@dave-laptop:~# python http.py Traceback (most recent call last): File "http.py", line 9, in conn.request("POST","http://202.45.139.58:8080/ppod-web",params,headers) File "/usr/lib/python2.6/httplib.py", line 898, in request self._send_request(method, url, body, headers) File "/usr/lib/python2.6/httplib.py", line 935, in _send_request self.endheaders() File "/usr/lib/python2.6/httplib.py", line 892, in endheaders self._send_output() File "/usr/lib/python2.6/httplib.py", line 764, in _send_output self.send(msg) File "/usr/lib/python2.6/httplib.py", line 723, in send self.connect() File "/usr/lib/python2.6/httplib.py", line 704, in connect self.timeout) File "/usr/lib/python2.6/socket.py", line 500, in create_connection for res in getaddrinfo(host, port, 0, SOCK_STREAM): socket.gaierror: [Errno -2] Name or service not known can anyone help me the code should be like this when rum in the url http://223.56.124.58:8080/wireless?data={"wifi":{"ssid":"guest","rssi","80"}} how to set like this or other way to do this to send to the server thanks

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  • python-social-auth AuthCanceled exception

    - by vero4ka
    I'm using python-social-auth in my Django application for authentication via Facebook. But when a user tries to login and when it's been refirected to Facebook app page clicks on "Cancel" button, appears the following exception: ERROR 2014-01-03 15:32:15,308 base :: Internal Server Error: /complete/facebook/ Traceback (most recent call last): File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/django/core/handlers/base.py", line 114, in get_response response = wrapped_callback(request, *callback_args, **callback_kwargs) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/django/views/decorators/csrf.py", line 57, in wrapped_view return view_func(*args, **kwargs) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/apps/django_app/utils.py", line 45, in wrapper return func(request, backend, *args, **kwargs) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/apps/django_app/views.py", line 21, in complete redirect_name=REDIRECT_FIELD_NAME, *args, **kwargs) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/actions.py", line 54, in do_complete *args, **kwargs) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/strategies/base.py", line 62, in complete return self.backend.auth_complete(*args, **kwargs) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/backends/facebook.py", line 63, in auth_complete self.process_error(self.data) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/backends/facebook.py", line 56, in process_error super(FacebookOAuth2, self).process_error(data) File "/home/vera/virtualenv/myapp/local/lib/python2.7/site-packages/social/backends/oauth.py", line 312, in process_error raise AuthCanceled(self, data.get('error_description', '')) AuthCanceled: Authentication process canceled Is the any way to catch it Django?

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  • Has Twisted changed its dependencies?

    - by cdecker
    Hi all, I'm currently working on a Python/Twisted project which is to be distributed and tested on Planetlab. For some reason my code was working on friday and now that I wanted to test a minor change it refuses to work at all: Traceback (most recent call last): File "acn_a4/src/node.py", line 6, in <module> from twisted.internet.protocol import DatagramProtocol File "/usr/lib/python2.5/site-packages/Twisted-10.0.0-py2.5-linux-i686.egg/twisted/__init__.py", line 18, in <module> from twisted.python import compat File "/usr/lib/python2.5/site-packages/Twisted-10.0.0-py2.5-linux-i686.egg/twisted/python/compat.py", line 146, in <module> import operator File "/home/cdecker/dev/acn/acn_a4/src/operator.py", line 7, in <module> File "/home/cdecker/acn_a4/src/node.py", line 6, in <module> from twisted.internet.protocol import DatagramProtocol File "/usr/lib/python2.5/site-packages/Twisted-10.0.0-py2.5-linux-i686.egg/twisted/internet/protocol.py", line 20, in <module> from twisted.python import log, failure, components File "/usr/lib/python2.5/site-packages/Twisted-10.0.0-py2.5-linux-i686.egg/twisted/python/log.py", line 19, in <module> from twisted.python import util, context, reflect File "/usr/lib/python2.5/site-packages/Twisted-10.0.0-py2.5-linux-i686.egg/twisted/python/util.py", line 5, in <module> import os, sys, hmac, errno, new, inspect, warnings File "/usr/lib/python2.5/inspect.py", line 32, in <module> from operator import attrgetter ImportError: cannot import name attrgetter And since I'm pretty new to python I have no idea what could have caused this problem. All suggestions are welcome :-)

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  • django: can't adapt error when importing data from postgres database

    - by Oleg Tarasenko
    Hi, I'm having strange error with installing fixture from dumped data. I am using psycopg2, and django1.1.1 silver:probsbox oleg$ python manage.py loaddata /Users/oleg/probs.json Installing json fixture '/Users/oleg/probs' from '/Users/oleg/probs'. Problem installing fixture '/Users/oleg/probs.json': Traceback (most recent call last): File "/opt/local/lib/python2.5/site-packages/django/core/management/commands/loaddata.py", line 153, in handle obj.save() File "/opt/local/lib/python2.5/site-packages/django/core/serializers/base.py", line 163, in save models.Model.save_base(self.object, raw=True) File "/opt/local/lib/python2.5/site-packages/django/db/models/base.py", line 495, in save_base result = manager._insert(values, return_id=update_pk) File "/opt/local/lib/python2.5/site-packages/django/db/models/manager.py", line 177, in _insert return insert_query(self.model, values, **kwargs) File "/opt/local/lib/python2.5/site-packages/django/db/models/query.py", line 1087, in insert_query return query.execute_sql(return_id) File "/opt/local/lib/python2.5/site-packages/django/db/models/sql/subqueries.py", line 320, in execute_sql cursor = super(InsertQuery, self).execute_sql(None) File "/opt/local/lib/python2.5/site-packages/django/db/models/sql/query.py", line 2369, in execute_sql cursor.execute(sql, params) File "/opt/local/lib/python2.5/site-packages/django/db/backends/util.py", line 19, in execute return self.cursor.execute(sql, params) ProgrammingError: can't adapt First I've checked similar issues on internet. This one seemed to be very related: http://code.djangoproject.com/ticket/5996, as my data has many non ASCII symbols But actually I've checked my django installation and it's ok there Could you advice what is wrong

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  • Django upload failing on request data read error

    - by Jake
    Hi All, I've got a Django app that accepts uploads from jQuery uploadify, a jQ plugin that uses flash to upload files and give a progress bar. Files under about 150k work, but bigger files always fail and almost always at around 192k (that's 3 chunks) completed, sometimes at around 160k. The Exception I get is below. exceptions.IOError request data read error File "/usr/lib/python2.4/site-packages/django/core/handlers/wsgi.py", line 171, in _get_post self._load_post_and_files() File "/usr/lib/python2.4/site-packages/django/core/handlers/wsgi.py", line 137, in _load_post_and_files self._post, self._files = self.parse_file_upload(self.META, self.environ[\'wsgi.input\']) File "/usr/lib/python2.4/site-packages/django/http/__init__.py", line 124, in parse_file_upload return parser.parse() File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 192, in parse for chunk in field_stream: File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 314, in next output = self._producer.next() File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 468, in next for bytes in stream: File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 314, in next output = self._producer.next() File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 375, in next data = self.flo.read(self.chunk_size) File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 405, in read return self._file.read(num_bytes) When running locally on the Django development server, big files work. I've tried setting my FILE_UPLOAD_HANDLERS = ("django.core.files.uploadhandler.TemporaryFileUploadHandler",) in case it was the memory upload handler, but it made no difference. Does anyone know how to fix this?

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  • How can I do such a typical unittest?

    - by Malcom.Z
    This is a simple structure in my project: MyAPP--- note--- __init__.py views.py urls.py test.py models.py auth-- ... template--- auth--- login.html register.html note--- noteshow.html media--- css--- ... js--- ... settings.py urls.py __init__.py manage.py I want to make a unittest which can test the noteshow page working propeyly or not. The code: from django.test import TestCase class Note(TestCase): def test_noteshow(self): response = self.client.get('/note/') self.assertEqual(response.status_code, 200) self.assertTemplateUsed(response, '/note/noteshow.html') The problem is that my project include an auth mod, it will force the unlogin user redirecting into the login.html page when they visit the noteshow.html. So, when I run my unittest, in the bash it raise an failure that the response.status_code is always 302 instead of 200. All right though through this result I can check the auth mod is running well, it is not like what I want it to be. OK, the question is that how can I make another unittest to check my noteshow.template is used or not? Thanks for all. django version: 1.1.1 python version: 2.6.4 Use Eclipse for MAC OS

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  • Python3 and ftplib uploading files

    - by Teifion
    My python2 script uploads files nicely using this method but python3 is presenting problems and I'm stuck as to where to go next (googling hasn't helped). from ftplib import FTP ftp = FTP(ftp_host, ftp_user, ftp_pass) ftp.storbinary('STOR myfile.txt', open('myfile.txt')) The error I get is Traceback (most recent call last): File "/Library/WebServer/CGI-Executables/rob3/functions/cli_f.py", line 12, in upload ftp.storlines('STOR myfile.txt', open('myfile.txt')) File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 454, in storbinary conn.sendall(buf) TypeError: must be bytes or buffer, not str I tried altering the code to from ftplib import FTP ftp = FTP(ftp_host, ftp_user, ftp_pass) ftp.storbinary('STOR myfile.txt'.encode('utf-8'), open('myfile.txt')) But instead I got this Traceback (most recent call last): File "/Library/WebServer/CGI-Executables/rob3/functions/cli_f.py", line 12, in upload ftp.storbinary('STOR myfile.txt'.encode('utf-8'), open('myfile.txt')) File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 450, in storbinary conn = self.transfercmd(cmd) File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 358, in transfercmd return self.ntransfercmd(cmd, rest)[0] File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 329, in ntransfercmd resp = self.sendcmd(cmd) File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 244, in sendcmd self.putcmd(cmd) File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 179, in putcmd self.putline(line) File "/Library/Frameworks/Python.framework/Versions/3.1/lib/python3.1/ftplib.py", line 172, in putline line = line + CRLF TypeError: can't concat bytes to str Can anybody point me in the right direction

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  • Orbited exception Data must not be unicode.

    - by Sid
    I am working with orbited and once I switch on orbited in production mode it throws the following error on my screen -- <exception caught here> --- File "/usr/lib/python2.6/dist-packages/twisted/web/server.py", line 150, in process self.render(resrc) File "/usr/lib/python2.6/dist-packages/twisted/web/server.py", line 157, in render body = resrc.render(self) File "/usr/local/lib/python2.6/dist-packages/orbited-0.7.10-py2.6.egg/orbited/transports/base.py", line 21, in render self.conn.transportOpened(self) File "/usr/local/lib/python2.6/dist-packages/orbited-0.7.10-py2.6.egg/orbited/cometsession.py", line 322, in transportOpened self.cometTransport.flush() File "/usr/local/lib/python2.6/dist-packages/orbited-0.7.10-py2.6.egg/orbited/transports/base.py", line 45, in flush self.write(self.packets) File "/usr/local/lib/python2.6/dist-packages/orbited-0.7.10-py2.6.egg/orbited/transports/htmlfile.py", line 42, in write self.request.write(payload); File "/usr/lib/python2.6/dist-packages/twisted/web/http.py", line 862, in write self.transport.write(data) File "/usr/lib/python2.6/dist-packages/twisted/internet/tcp.py", line 420, in write abstract.FileDescriptor.write(self, bytes) File "/usr/lib/python2.6/dist-packages/twisted/internet/abstract.py", line 170, in write raise TypeError("Data must not be unicode") exceptions.TypeError: Data must not be unicode I have absolutely no clue as to what could be the problem. Could anyone point me in the right direction.

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  • tracd server problems

    - by deddihp
    Hello, I got the following error while accessing tracd server, what's going on ? Thanks. [oke@localhost Trac-0.11.7]$ sudo tracd -p 8000 /home/deddihp/trac/ Server starting in PID 5082. Serving on 0.0.0.0:8000 view at http://127.0.0.1:8000/ ---------------------------------------- Exception happened during processing of request from ('127.0.0.1', 47804) Traceback (most recent call last): File "/usr/lib/python2.6/SocketServer.py", line 558, in process_request_thread self.finish_request(request, client_address) File "/usr/lib/python2.6/SocketServer.py", line 320, in finish_request self.RequestHandlerClass(request, client_address, self) File "/usr/lib/python2.6/SocketServer.py", line 615, in __init__ self.handle() File "/usr/lib/python2.6/BaseHTTPServer.py", line 329, in handle self.handle_one_request() File "/usr/lib/python2.6/site-packages/Trac-0.11.7-py2.6.egg/trac/web/wsgi.py", line 194, in handle_one_request gateway.run(self.server.application) File "/usr/lib/python2.6/site-packages/Trac-0.11.7-py2.6.egg/trac/web/wsgi.py", line 94, in run response = application(self.environ, self._start_response) File "/usr/lib/python2.6/site-packages/Trac-0.11.7-py2.6.egg/trac/web/standalone.py", line 100, in __call__ return self.application(environ, start_response) File "/usr/lib/python2.6/site-packages/Trac-0.11.7-py2.6.egg/trac/web/main.py", line 346, in dispatch_request locale.setlocale(locale.LC_ALL, environ['trac.locale']) File "/usr/lib/python2.6/locale.py", line 513, in setlocale return _setlocale(category, locale) Error: unsupported locale setting ----------------------------------------

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  • Python: eliminating stack traces into library code?

    - by Mark Harrison
    When I get a runtime exception from the standard library, it's almost always a problem in my code and not in the library code. Is there a way to truncate the exception stack trace so that it doesn't show the guts of the library package? For example, I would like to get this: Traceback (most recent call last): File "./lmd3-mkhead.py", line 71, in <module> main() File "./lmd3-mkhead.py", line 66, in main create() File "./lmd3-mkhead.py", line 41, in create headver1[depotFile]=rev TypeError: Data values must be of type string or None. and not this: Traceback (most recent call last): File "./lmd3-mkhead.py", line 71, in <module> main() File "./lmd3-mkhead.py", line 66, in main create() File "./lmd3-mkhead.py", line 41, in create headver1[depotFile]=rev File "/usr/anim/modsquad/oses/fc11/lib/python2.6/bsddb/__init__.py", line 276, in __setitem__ _DeadlockWrap(wrapF) # self.db[key] = value File "/usr/anim/modsquad/oses/fc11/lib/python2.6/bsddb/dbutils.py", line 68, in DeadlockWrap return function(*_args, **_kwargs) File "/usr/anim/modsquad/oses/fc11/lib/python2.6/bsddb/__init__.py", line 275, in wrapF self.db[key] = value TypeError: Data values must be of type string or None.

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  • Configure Django project in a subdirectory using mod_python. Admin not working.

    - by David
    HI guys. I was trying to configure my django project in a subdirectory of the root, but didn't get things working.(LOcally it works perfect). I followed the django official django documentarion to deploy a project with mod_python. The real problem is that I am getting "Page not found" errors, whenever I try to go to the admin or any view of my apps. Here is my python.conf file located in /etc/httpd/conf.d/ in Fedora 7 LoadModule python_module modules/mod_python.so SetHandler python-program PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE mysite.settings PythonOption django.root /mysite PythonDebug On PythonPath "['/var/www/vhosts/mysite.com/httpdocs','/var/www/vhosts/mysite.com/httpdocs/mysite'] + sys.path" I know /var/www/ is not the best place to put my django project, but I just want to send a demo of my work in progress to my customer, later I will change the location. For example. If I go to www.domain.com/mysite/ I get the index view I configured in mysite.urls. But I cannot access to my app.urls (www.domain.com/mysite/app/) and any of the admin.urls.(www.domain.com/mysite/admin/) Here is mysite.urls: urlpatterns = patterns('', url(r'^admin/password_reset/$', 'django.contrib.auth.views.password_reset', name='password_reset'), (r'^password_reset/done/$', 'django.contrib.auth.views.password_reset_done'), (r'^reset/(?P<uidb36>[0-9A-Za-z]+)-(?P<token>.+)/$', 'django.contrib.auth.views.password_reset_confirm'), (r'^reset/done/$', 'django.contrib.auth.views.password_reset_complete'), (r'^$', 'app.views.index'), (r'^admin/', include(admin.site.urls)), (r'^app/', include('mysite.app.urls')), (r'^photologue/', include('photologue.urls')), ) I also tried changing admin.site.urls with ''django.contrib.admin.urls' , but it didn't worked. I googled a lot to solve this problem and read how other developers configure their django project, but didn't find too much information to deploy django in a subdirectory. I have the admin enabled in INSTALLED_APPS and the settings.py is ok. Please if you have any guide or telling me what I am doing wrong it will be much appreciated. THanks.

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  • Testing variable types in Python

    - by Jasper
    Hello, I'm creating an initialising function for the class 'Room', and found that the program wouldn't accept the tests I was doing on the input variables. Why is this? def __init__(self, code, name, type, size, description, objects, exits): self.code = code self.name = name self.type = type self.size = size self.description = description self.objects = objects self.exits = exits #Check for input errors: if type(self.code) != type(str()): print 'Error found in module rooms.py!' print 'Error number: 110' elif type(self.name) != type(str()): print 'Error found in module rooms.py!' print 'Error number: 111' elif type(self.type) != type(str()): print 'Error found in module rooms.py!' print 'Error number: 112' elif type(self.size) != type(int()): print 'Error found in module rooms.py!' print 'Error number: 113' elif type(self.description) != type(str()): print 'Error found in module rooms.py!' print 'Error number: 114' elif type(self.objects) != type(list()): print 'Error found in module rooms.py!' print 'Error number: 115' elif type(self.exits) != type(tuple()): print 'Error found in module rooms.py!' print 'Error number: 116' When I run this I get this error: Traceback (most recent call last): File "/Users/Jasper/Development/Programming/MyProjects/Game Making Challenge/Europa I/rooms.py", line 148, in <module> myRoom = Room(101, 'myRoom', 'Basic Room', 5, '<insert description>', myObjects, myExits) File "/Users/Jasper/Development/Programming/MyProjects/Game Making Challenge/Europa I/rooms.py", line 29, in __init__ if type(self.code) != type(str()): TypeError: 'str' object is not callable

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  • ubuntu 13.10 kvm binary is deprecated, please use qemu-system-x86_64

    - by ??1986
    I just upgrade from 13.04 to 13.10 and I have this issue when I run my KVM Unable to complete install: 'internal error: process exited while connecting to monitor: W: kvm binary is deprecated, please use qemu-system-x86_64 instead char device redirected to /dev/pts/10 (label charserial0) failed to initialize KVM: Device or resource busy Detail Error: Traceback (most recent call last): File "/usr/share/virt-manager/virtManager/asyncjob.py", line 96, in cb_wrapper callback(asyncjob, *args, **kwargs) File "/usr/share/virt-manager/virtManager/create.py", line 1983, in do_install guest.start_install(False, meter=meter) File "/usr/lib/python2.7/dist-packages/virtinst/Guest.py", line 1246, in start_install noboot) File "/usr/lib/python2.7/dist-packages/virtinst/Guest.py", line 1314, in _create_guest dom = self.conn.createLinux(start_xml or final_xml, 0) File "/usr/lib/python2.7/dist-packages/libvirt.py", line 2892, in createLinux if ret is None:raise libvirtError('virDomainCreateLinux() failed', conn=self) libvirtError: internal error: process exited while connecting to monitor: W: kvm binary is deprecated, please use qemu-system-x86_64 instead char device redirected to /dev/pts/8 (label charserial0) failed to initialize KVM: Device or resource busy

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  • Am I correctly handling duplicate URLs for my homepage?

    - by Rob Goldstein
    I own a Job Search site named www.conservationjobboard.com and have a concern about how the domain is viewed by search engines. The issue is that when the site was first designed, the default page was left as default.php, but the homepage was actually JobBoard.php. To handle this, the default.php page performed a redirect to the JobBoard.php file when www.conservationjobboard.com/ was requested. The main problem resulted because the redirect was a temporary redirect causing search engines to index conservationjobboard.com/ and conservationjobboard.com/JobBoard.php as 2 separate pages. This has since been corrected to use the .htaccess file so that JobBoard.php is now the default file for the root directory eliminating the need for the redirect. Problem is that search engines still show both URL's in search results (one including JobBoard.php and one that ends with /). Another potential problem is that some of my early backlinks are to conservationjobboard.com/JobBoard.php while the rest are to conservationjobboard.com The 2 outstanding questions are as follows: 1. Is my domain still being penalized by search engines like Google for having duplicate homepage URL's? 2. Are all of the back links to my homepage being considered as the same now or is the total number of back links being split between the 2 different URL's? If you think there are still issues with how we have this set-up, I was wondering if you could give me advice on what we should do differently. Thanks.

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  • Why do some user agents have spam urls in them?

    - by Erx_VB.NExT.Coder
    If you go to (say) the last 100 entries (visits) to the botsvsbrowsers.com website (exact link, feel free to take a look: http://www.botsvsbrowsers.com/recent/listings/index.html ), you'd notice that almost every User Agent that has the keywords "Opera" and "Presto" inside them, will almost certainly have a web link (URL/Web Address) inside it, and it won't just be a normal web address, but a HTML anchor tag/link to that address. Why is this so, I could not even find a single discussion about it on the internet, nowhere, I tried varying my search terms many times. If the user agent contains the words "Opera" and "Presto" it doesnt mean it will have this weblink, but it means there is about an 80% change that it will. A typical anchor tag/link inside a user agent will look like this: Mozilla/4.0 <a href="http://osis-uk.co.uk/disabled-equipment">disability equipment</a> (Windows NT 5.1; U; en) Presto/2.10.229 Version/11.60 If you check it out at the website, http://www.botsvsbrowsers.com/recent/listings/index.html you will notice that the back and forward arrows are in there unescaped format. This isn't just true for botsvsbrowsers, but several other user agent listing sites. I'm really confused and feel line I'm in a room full of 10,000 people and am the only one seeing this ghost :). If I'm doing statistical analysis, should I include or exclude this type of user agent from my listing (ie: are these just normal users who've set their user agents to attempt to drive some traffic to their sites as they browser the web), or is there something else going on? The fact that it is so consistent in terms of its format leads me to believe that it is an automated process (the setting or alteration of the user agent) so I cannot decide or understand the process by which this change is made (I know how to change a user agent), but unsure which program or facility is doing this, especially since it is exclusive to Opera (Presto) user agents that are beyond I think an 8 or 9 point something browser version. I've run some statistical tests, parsing entries from all over the place, writing custom programs, to get a better understanding of this. Keep in mind that I see normal URL's in user agents infrequently, they are just text such as +http://www.someSite.com appended to a user agent normally, especially if its a crawler or bot it provided its service URL, this is normal and isnt done with an embedded link (A HREF=) etc, so I'm not talking about "those".

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  • problem using pydoc in python

    - by rohanag
    I'm using pydoc in python 2.7.3 to generate documentation for a file called PreProcessingAPI.py which contains a class called PreProcessingAPI In PreProcessingAPI.py, I have the following import in the beginning of the file: from __future__ import division from re import * from nltk.stem import porter The problem is, in the documentation generated by pydoc, nltk.stem.porter is shown as a Module. There is also a DATA heading with all sorts of variables I do not know about. Is there a way to avoid these variables and avoid showing nltk.stem.porter in the modules? I'm running the following command to generate documentation python pydoc.py -w PreProcessingAPI.py I've put the file pydoc.py in the directory containing my file. Here is the file generated: https://www.dropbox.com/s/4rb6ut99o25mwly/PreProcessingAPI.html

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  • for a blog with posts and categories what are all the best ways to create user friendly and seo friendly urls

    - by Jayapal Chandran
    I am creating a module in my website which displays ringtones. it is like creating blog posts and categories It will have categories(tags) and posts. (i am using category and tag interchangeably) i am using the following linking for this module sitename.com/blog sitename.com/blog/category/category-name-slug/ - will list all ringtones of that category/tag sitename.com/blog/title/name-slug-of-the-ringtone/ - this will display the details and a download link in all page at the left i display the category/tag . This is how i have formed the url structure. it will be user friendly i hope yet will it be seo friendly? Please hint if i am missing something or other ways to improve. meanwhile i am browsing the net to get more information on linking content (categorizing) and to find best ways for the user and search engine.

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  • Why is a # sign added to the end of URLS?

    - by Niro
    Note: I'm asking this from the perspective of the site developers (trying to help someone there). not as a user. Please don't forward this to superuser.com. It's a server admin question. Have a look here http://www.wanimo.com/fr/chiens/coussin-matelas-tapis-pour-chien-sc28/tapis-plat-urban-chic-sf7263/ you'll see that the page gets redirected to the same page with # at the end. Worse, when you click back you get garbage url. I'm trying to debug what is causing the redirect. Any advice on how to find it ?

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  • Should I add a "nofollow" attribute to download links, or disallow the URLs in robots.txt?

    - by Laurent
    I have a download link very similar to Opera's one - it's just a script that sends the file. It doesn't have an extension and there's no obvious way to tell that it's actually a download link. So since I don't want robots to crawl this link, do I need to add it to robots.txt or maybe add a "nofollow" attribute to it? I see that on Opera's website they didn't do either of this, so perhaps it's not necessary?

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  • What should filenames and URLs of images contain for SEO benefit?

    - by Baumr
    We know that good site architecture usually looks like this: example-company.com/ example-company.com/about/ example-company.com/contact/ example-company.com/products/ example-company.com/products/category/ example-company.com/products/category/productname/ Now, when it comes to Google Image search, it is clear that the img alt tag, filename/URL, and surrounding text (captions, headings, paragraphs) have an effect on ranking. I want to ask about the filename of the images that we should use (e.g. product-photo.jpg). ...but first about the URL: Often web developers stick all images in a single folder in the root: example-company.com/img/ — and I have stopped doing that. (I don't want to get into it, but basically, it seems more semantic for images which make up part of the content at each sub-directory) However, when all images appear in a folder, I feel that their filename needs to reflect what they are a bit more than usual, for example: example-company.com/img/example-company-productname-category.jpg It's a longer filename than just product.png, but as long as it's relevant, I see no problem with regards to SEO (unless you're keyword stuffing), and it could even help rank for keywords: "example company" "productname" "category" So no questions there. But what about when we have places images in the site architecture we outlined at the beginning? In other words, what if image URL paths look like this: example-company.com/products/category/productname/productname.jpg My question is, should the URL be kept short like above and only have the "productname" (and some descriptive keywords) as part of it's filename? Or, should it also include the "example-company" and "category"? Like so: example-company.com/products/category/productname/example-company-category-productname.jpg That seems much longer, and redundant when we look at the URL, but here are a few considerations. Images are often downloaded onto computers, and, to the average user, they lose their original URL and thus — it isn't clear where they came from. Also, some social networks, forums, and other platforms leave the filename intact when uploaded. (Many others rewrite it, for example, Pinterest and Facebook.) Another consideration, will this really help (even if ever so slightly) rank in Google Image Search, or at least inform Google that the product is something specific to the "example-company"? For example, what if this product can only be bought at this store and is the flagship product? In addition to an abundance of internal links to this product page, would having the "example company" name and "category" help it appear in "example company" searches? In other words, is less more?

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  • What's the best way to version CSS and JS URLs?

    - by David Eyk
    As per Yahoo's much-ballyhooed Best Practices for Speeding Up Your Site, we serve up static content from a CDN using far-future cache expiration headers. Of course, we need to occasionally update these "static" files, so we currently add an infix version as part of the filename (based on the SHA1 sum of the file contents). Thus: styles.min.css Becomes: styles.min.abcd1234.css However, managing the versioned files can become tedious, and I was wondering if a GET argument notation might be cleaner and better: styles.min.css?v=abcd1234 Which do you use, and why? Are there browser- or proxy/cache-related considerations that I should consider?

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  • How can I test for a URLs existeance before redirecting to it?

    - by ckliborn
    I am using Apache's mod_rewrite to redirect mobile users to my mobile site based on their http_user_agent. However not all pages have a mobile equivalent. Also mobile pages end in .html and "full" pages end in .shtml. Here is some pseudo code. Does the user have a certain HTTP_USER_AGENT? Is there a mobile page? If so take them there. If not, no redirection is needed. I want to do this with apache.

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  • .htaccess rules to rewrite URLs to front end page?

    - by Dizzley
    I am adding a new application to my site at example.com/app. I want views at that URL to always open myapp.php. E.g. example.com/app -> example.com/app/myapp.php and example.com/app/ -> example.com/app/myapp.php What's the correct form of rewrite rules in the .htaccess file? I've got: <IfModule mod_rewrite.c> RewriteEngine On RewriteBase /app/ RewriteRule ^myapp\.php$ - [L] RewriteRule ^myapp.php$ - [L] RewriteRule . - [L] </IfModule> ...based on what the Wordpress front-end does. But all I see at example.com/app is a directory of files. :( (I put those rewrites at the top of my .htaccess file). Any ideas? Update What actually worked: RewriteEngine On RewriteBase / RewriteCond %{REQUEST_URI} ^/app(/.*)?$ [NC] RewriteCond %{REQUEST_FILENAME} !-f RewriteRule . /app/myapp.php [L] This is good because: Explicit or implicit calls to app/myapp.php work. example.com/app redirects to app/myapp.php example.com/app/ redirects to app/myapp.php example.com/app/subfunction redirects to app/myapp.php All other calls to example.com/otherstuff are untouched. Item 4 is Wordpress-like Front Controller pattern behaviour. I think that rule RewriteCond %{REQUEST_URI} ^/app.*$ [NC] needs refining as it allows /app-oh-my-goodness etc. through too. Thanks for the answers.

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