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  • Blocking error in celery

    - by dmitry
    I have no idea what's this. Python 2.7 + django-1.5.1 + httpd + rabbitmq + django-celery==3.0.17 Tasks are not executed because of some error. Below is celery's log. Maybe someone has faced it before. [2013-06-24 17:10:03,792: CRITICAL/MainProcess] Can't decode message body: AttributeError("'JoinInfo' object has no attribute '__dict__'",) (type:u'application/x-python-serialize' encoding:u'binary' raw:'\'\\x80\\x02}q\\x01(U\\x07expiresq\\x02NU\\x03utcq\\x03\\x88U\\x04argsq\\x04cdjango.contrib.auth.models\\nUser\\nq\\x05)\\x81q\\x06}q\\x07(U\\x08usernameq\\x08X\\x19\\x00\\x00\\[email protected]\\nfirst_nameq\\tX\\x05\\x00\\x00\\x00BibbyU\\tlast_nameq\\nX\\x08\\x00\\x00\\x00OffshoreU\\r_client_cacheq\\x0bccopy_reg\\n_reconstructor\\nq\\x0ccbongoregistration.models\\nClient\\nq\\rc__builtin__\\nobject\\nq\\x0eN\\x87Rq\\x0f}q\\x10(h\\nX\\x08\\x00\\x00\\x00OffshoreU\\x1bpurchase_confirmation_emailq\\x11X\\x1f\\x00\\x00\\[email protected]\\x1dpurchase_confirmation_email_1q\\x12X!\\x00\\x00\\[email protected]\\x06_stateq\\x13cdjango.db.models.base\\nModelState\\nq\\x14)\\x81q\\x15}q\\x16(U\\x06addingq\\x17\\x89U\\x02dbq\\x18U\\x07defaultq\\x19ubU\\x0buser_ptr_idq\\x1aJ\\xb4\\xa2\\x03\\x00U\\x08is_staffq\\x1b\\x89U\\x08postcodeq\\x1cX\\x08\\x00\\x00\\x00AB11 5BSU\\x0cdegree_limitq\\x1dK\\x06U\\x07messageq\\x1eX\\xd1E\\x00\\x00<table id="container" style="margin: 0px; padding: 0px; width: 100%; background-color: #ffffff;" cellspacing="0" cellpadding="0"... (22911b)'') Traceback (most recent call last): File "/opt/www/MyProject-main/eggs/kombu-2.5.10-py2.7.egg/kombu/messaging.py", line 556, in _receive_callback decoded = None if on_m else message.decode() File "/opt/www/MyProject-main/eggs/kombu-2.5.10-py2.7.egg/kombu/transport/base.py", line 147, in decode self.content_encoding, accept=self.accept) File "/opt/www/MyProject-main/eggs/kombu-2.5.10-py2.7.egg/kombu/serialization.py", line 187, in decode return decode(data) File "/opt/www/MyProject-main/eggs/kombu-2.5.10-py2.7.egg/kombu/serialization.py", line 74, in pickle_loads return load(BytesIO(s)) AttributeError: 'JoinInfo' object has no attribute '__dict__'

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  • Firefox handles xxx.submit(), Safari doesn't ... what can be done?

    - by Prairiedogg
    I'm trying to make a pull down menu post a form when the user selects (releases the mouse) on one of the options from the menu. This code works fine in FF but Safari, for some reason, doesn't submit the form. I re-wrote the code using jquery to see if jquery's .submit() implementation handled the browser quirks better. Same result, works in FF doesn't work in safari. The following snippets are from the same page, which has some django template language mixed in. Here's the vanilla js attempt: function formSubmit(lang) { if (lang != '{{ LANGUAGE_CODE }}') { document.getElementById("setlang_form").submit(); } } Here's the jquery attempt: $(document).ready(function() { $('#lang_submit').hide() $('#setlang_form option').mouseup(function () { if ($(this).attr('value') != '{{ LANGUAGE_CODE }}') { $('#setlang_form').submit() } }); }); and here's the form: <form id="setlang_form" method="post" action="{% url django.views.i18n.set_language %}"> <fieldset> <select name="language"> {% for lang in interface_languages %} <option value="{{ lang.code }}" onmouseup="formSubmit('{{ lang.name }}')" {% ifequal lang.code LANGUAGE_CODE %}selected="selected"{% endifequal %}>{{ lang.name }}</option> {% endfor %} </select> </fieldset> </form> My question is, how can I get this working in Safari?

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  • Nginx rewrite with Simple Machines Forum

    - by Kevin Worthington
    I am running Nginx 1.5.6 and I use the Simple Machines Forum software. Most rewrite rules seem to work properly, with the exception of the RSS feeds. In my Nginx configuration, I have the following line which is supposed to handle URLs which contain ".xml": rewrite ^/forum/(\.xml|xmlhttp)/?$ "/forum/index.php?pretty;action=$1" last; The above rule produces the following URL for the main forum, which returns a 403 Error: http://www.mydomain.com/forum/.xml/?type=rss I would like the rewrite rule to produce this type of URL, which returns code 200 (a real page): http://www.mydomain.com/forum/?type=rss;action=.xml Here is the entire block pertaining to the forum rewrites: http://pastebin.com/raw.php?i=tZkAibW3 I would really appreciate some help to create a rewrite rule to do that. Thanks.

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  • Setting up podcasting for a non-tech user

    - by Force Flow
    I have a user who wants to start making podcasts, but they only have basic skills when it comes to technology. So, I was trying to get a process together that would be easy for them to follow. To upload files (the mp3's and rss feed files), I have an explorer shortcut for their FTP space. To record the podcast, I was going to either use audacity or PodProducer. For the RSS feed, I was looking for a podcast RSS generator of some sort. In my search for this, I've come across a lot of dead links and a lot of paid tools, so I haven't come up with anything too useful. Is there a free, reliable webservice or windows-based tool available that folks like to use?

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  • Is it possible to create a service like Feed My Inbox on my own server?

    - by Mark Bowen
    I was just wondering if it's at all possible to create a service like Feed My Inbox on my own server using PHP? Basically I have a site which has RSS feeds which are dynamic in nature and can search from thousands of posts based on many different criteria. I have the RSS feed working fine and bringing back data dynamically for whatever criteria I want so that bits fine. I am using the ExpressionEngine CMS to handle the site and there will be thousands of users on the site (currently there are around 2,0000) but that number is exponentially growing every single day. What I want to be able to do is allow the users to choose from certain criteria which will then build a dynamic RSS URL which will then be stored in a database table (one row for each user). This bit I will be able to do myself but then I want to be able to send out new RSS feed items via e-mail to each user. This is the part I'm a little stuck on. I'm guessing I would somehow need to run a cron job to hit a page which would check each users RSS feed and then if there are new items to send them to the user via e-mail. That's where I am totally stuck though and I'm just wondering what the best way to go about it would be? That or any software in PHP that already does this sort of thing would be great. I tried out phpList but it has severe problems working with RSS and I only ever got it to work once and now never again and I've read that lots of people have had this same problem so unfortunately it's not just me :-( I know there are services such as Feed My Inbox which I could easily set up so that users click a link and their RSS feed URL is added to go and use that service but I want to keep users from seeing the dynamic nature of the feed or they will easily be able to modify it to get at other items in the feed. I need this so that I can charge for access to the feeds but if people can see the URL of the feed then I will be totally unstuck as they will be able to get at whatever they want very easily. Therefore I'd like to be able to send the items out to them. Would really love to hear if anyone knows if this kind of thing is possible at all and what would be involved?

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  • Generic relations missing with grappelli

    - by diegueus9
    I'm using the last svn revision of grappelli and rev 11840 of django (before multidatabases and stuff), and i'm trying to use generic relations in the admin, but doesn't work, The model: class AutorProyectoLey(DatedModel): tipo_autor = models.ForeignKey(ContentType) autor_id = models.PositiveIntegerField() content_object = generic.GenericForeignKey('tipo_autor', 'autor_id') proyecto_ley = models.ForeignKey(ProyectoLey) The admin: class AutorInline(GenericInlineModelAdmin): model = AutorProyectoLey allow_add = True ct_field = 'tipo_autor' ct_fk_field = 'autor_id' classes = ('collapse-open',) And i put this model of var inlines in another adminmodel, but the html render is : <!-- Inlines --> <!-- Submit-Row -->

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  • 2 fields unique

    - by xRobot
    I have this model: class blog(models.Model): user = models.ForeignKey(User) mail = models.EmailField(max_length=60, null=False, blank=False) name = models.CharField(max_length=60, blank=True, null=True) I want that (user,email) are unique togheter. For example: This is allowed: 1, [email protected], myblog 2, [email protected], secondblog This is NOT allowed: 1, [email protected], myblog 1, [email protected], secondblog Is this possible in Django ?

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  • why save_model method doesn't work in admin.StackedInline?

    - by FurtiveFelon
    Hi all, I have a similar problem as a previously solved problem of mine, except this time solution doesn't seem to work: http://stackoverflow.com/questions/2991365/how-to-auto-insert-the-current-user-when-creating-an-object-in-django-admin Previously i used to override the save_model to stamp the user submitting the article. Now i need to do the same for comments, it doesn't seem to work anymore. Anyone have any ideas? Thanks a lot! Jason

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  • How to use the same template for different query sets?

    - by knuckfubuck
    I'm new to Django and setting up my first site. I have a Share model and a template called share_list.html that uses an object_list like this: {% for object in object_list %} I setup haystack using their tutorial and the search template looks like this: {% for result in page.object_list %} I would like to modify the search.html template to have an include of the share_list so I don't have to repeat myself. How can I make it use the same object_list?

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  • regular expression with special chars

    - by xRobot
    I need a regular expression to validate string with one or more of these characters: a-z A-Z ' àòèéùì simple white space FOR EXAMPLE these string are valide: D' argon calabrò maryòn l' Ancol these string are NOT valide: hello38239 my_house work [tab] with me I tryed this: re.match(r"^[a-zA-Z 'òàèéìù]+$", self.cleaned_data['title'].strip()) It seems to work in my python shell but in Django I get this error: SyntaxError at /home/ ("Non-ASCII character '\\xc3' ... Why ?

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  • apache solr auto suggestions

    - by Pydev UA
    I use solr+django-haystack I set settings.HAYSTACK_INCLUDE_SPELLING = True and rebuild index I'm trying to get any suggestion using: SearchQuerySet().auto_query('tryng ani word her').spelling_suggestion() But I always get None What should I do to get at least one working suggestion ? may be I need add some configuration into solr config or have some specific data indexed ?

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  • Overriding initial value in ModelForm

    - by schneck
    Hi, in my Django (1.2) project, I want to prepopulate a field in a modelform, but my new value is ignored. This is the snippet: class ArtefactForm(ModelForm): material = CharField(widget=AutoCompleteWidget('material', force_selection=False)) def __init__(self, *args, **kwargs): super(ArtefactForm, self).__init__(*args, **kwargs) self.fields['material'].initial = 'Test' I also tried with self.base_fields, but no effect: there is always the database-value displaying in the form. Any ideas?

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  • PHP Frameworks (CodeIgnitor, Yii, CakePHP) vs. Django

    - by niting
    I have to develop a site which has to accomodate around 2000 users a day and speed is a criterion for it. Moreover, the site is a user oriented one where the user will be able to log in and check his profile, register for specific events he/she wants to participate in. The site is to be hosted on a VPS server.Although I have pretty good experience with python and PHP but I have no idea how to use either of the framework. We have plenty of time to experiment and learn one of the above frameworks.Could you please specify which one would be preferred for such a scenario considering speed, features, and security of the site. Thanks, niting

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  • List display names from django models

    - by Ed
    I have an object: POP_CULTURE_TYPES = ( ('SG','Song'), ('MV', 'Movie'), ('GM', 'Game'), ('TV', 'TV'), ) class Pop_Culture(models.Model): name = models.CharField(max_length=30, unique=True) type = models.CharField(max_length=2, choices = POP_CULTURE_TYPES, blank=True, null=True) Then I have a function: def choice_list(request, modelname, field_name): mdlnm = get.model('mdb', modelname.lower()) mdlnm = mdlnm.objects.values_list(field_name, flat=True).distinct().order_by(field_name) return render_to_response("choice_list.html", { 'model' : modelname, 'field' : field_name, 'field_list' : mdlnm }) This gives me a distinct list of all the "type" entries in the database in the "field_list" variable passed in render_to_response. But I don't want a list that shows: SG MV I want a list that shows: Song Movie I can do this on an individual object basis if I was in the template object.get_type_display But how do I get a list of all of the unique "type" entries in the database as their full names for output into a template? I hope this question was clearly described. . .

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  • Best way to re-use the same django models and admin for multiple apps

    - by kepioo
    Given a reference app ( called guide), how can I create additional apps that will reuse the same model/admin/views than guide - the motivation behind is to be able to individually control each subapp. guide guideApp1 exact same models/admin/views than guide guideApp2 exact same models/admin/views than guide in the Admin site, I should have : 1 section for guideApp1 with all the tables defined in guide, that applies to guideApp1 1 section for guideApp12 with all the tables defined in guide, that applies to guideApp2

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  • Validating ModelChoiceField in Django forms

    - by Andrey
    I'm trying to validate a form containing a ModelChoiceField: state = forms.ModelChoiceField(queryset=State.objects.all(), empty_label=None) When it is used in normal circumstances, everything goes just fine. But I'd like to protect the form from the invalid input. It's pretty obvious that I must get forms.ValidationError when I put invalid value in this field, isn't it? But if I try to submit a form with a value 'invalid' in 'state' field, I get ValueError: invalid literal for int() with base 10: 'invalid' and not the expected forms.ValidationError. What should I do? I tried to place a def clean_state(self) to check this field but that didn't work plus I don't think this is a good solution, there must be something more simple but I just missed that.

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  • Generate unique hashes for django models

    - by becomingGuru
    I want to use unique hashes for each model rather than ids. I implemented the following function to use it across the board easily. import random,hashlib from base64 import urlsafe_b64encode def set_unique_random_value(model_object,field_name='hash_uuid',length=5,use_sha=True,urlencode=False): while 1: uuid_number = str(random.random())[2:] uuid = hashlib.sha256(uuid_number).hexdigest() if use_sha else uuid_number uuid = uuid[:length] if urlencode: uuid = urlsafe_b64encode(uuid)[:-1] hash_id_dict = {field_name:uuid} try: model_object.__class__.objects.get(**hash_id_dict) except model_object.__class__.DoesNotExist: setattr(model_object,field_name,uuid) return I'm seeking feedback, how else could I do it? How can I improve it? What is good bad and ugly about it?

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  • GAE/Django Templates (0.96) filters to get LENGTH of GqlQuery and filter it

    - by Halst
    I pass the query with comments to my template: COMM = CommentModel.gql("ORDER BY created") doRender(self,CP.template,{'CP':CP,'COMM':COMM, 'authorize':authorize()}) And I want to output the number of comments as a result, and I try to do things like that: <a href="...">{{ COMM|length }} comments</a> Thats does not work (yeah, since COMM is GqlQuery, not a list). What can I do with that? Is there a way to convert GqlQuery to list or is there another solution? (first question) Second question is, how to filter this list in template? Is there a construct like this: <a href="...">{{ COMM|where(reference=smth)|length }} comments</a> so that I can get not only the number of all comments, but only comments with certain db.ReferenceProperty() property, for example. Last question: is it weird to do such things using templates?

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  • Python Django MySQLdb setup problem:: setup.py dosen't build due to incorrect location of mysql

    - by 108860375137931889948
    I'm trying to install MySQLdb for python. but when I run the setup, this is the error I get. well I know why its giving all the missing file statements, but dont know where to change the bold marked location from. Please help gaurav-toshniwals-macbook-7:MySQL-python-1.2.3c1 gauravtoshniwal$ python setup.py build running build running build_py copying MySQLdb/release.py - build/lib.macosx-10.3-fat-2.6/MySQLdb running build_ext building '_mysql' extension gcc-4.0 -arch ppc -arch i386 -isysroot /Developer/SDKs/MacOSX10.4u.sdk -fno-strict-aliasing -fno-common -dynamic -DNDEBUG -g -O3 -Dversion_info=(1,2,3,'gamma',1) -D_version_=1.2.3c1 -I/Applications/MAMP/Library/include/mysql -I/Library/Frameworks/Python.framework/Versions/2.6/include/python2.6 -c _mysql.c -o build/temp.macosx-10.3-fat-2.6/_mysql.o _mysql.c:36:23: error: my_config.h: No such file or directory _mysql.c:36:23: error: my_config.h: No such file or directory _mysql.c:38:19: error: mysql.h: No such file or directory _mysql.c:38:19:_mysql.c:39:26: error: mysqld_error.h: No such file or directory error: _mysql.c:40:20:mysql.h: No such file or directory

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  • How to manage Javascript modules in django templates?

    - by John Mee
    Lets say we want a library of javascript-based pieces of functionality (I'm thinking jquery): For example: an ajax dialog a date picker a form validator a sliding menu bar an accordian thingy There are four pieces of code for each: some Python, CSS, JS, & HTML. What is the best way to arrange all these pieces so that: each javascript 'module' can be neatly reused by different views the four bits of code that make up the completed function stay together the css/js/html parts appear in their correct places in the response common dependencies between modules are not repeated (eg: a javascript file in common) x-------------- It would be nice if, or is there some way to ensure that, when called from a templatetag, the templates respected the {% block %} directives. Thus one could create a single template with a block each for CSS, HTML, and JS, in a single file. Invoke that via a templatetag which is called from the template of whichever view wants it. That make any sense. Can that be done some way already? My templatetag templates seem to ignore the {% block %} directives. x-------------- There's some very relevant gasbagging about putting such media in forms here http://docs.djangoproject.com/en/dev/topics/forms/media/ which probably apply to the form validator and date picker examples.

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  • ContentType Issue -- Human is an idiot - Can't figure out how to tie the original model to a Content

    - by bmelton
    Originally started here: http://stackoverflow.com/questions/2650181/django-in-query-as-a-string-result-invalid-literal-for-int-with-base-10 I have a number of apps within my site, currently working with a simple "Blog" app. I have developed a 'Favorite' app, easily enough, that leverages the ContentType framework in Django to allow me to have a 'favorite' of any type... trying to go the other way, however, I don't know what I'm doing, and can't find any examples for. I'll start off with the favorite model: favorite/models.py from django.db import models from django.contrib.contenttypes.models import ContentType from django.contrib.contenttypes import generic from django.contrib.auth.models import User class Favorite(models.Model): content_type = models.ForeignKey(ContentType) object_id = models.PositiveIntegerField() user = models.ForeignKey(User) content_object = generic.GenericForeignKey() class Admin: list_display = ('key', 'id', 'user') class Meta: unique_together = ("content_type", "object_id", "user") Now, that allows me to loop through the favorites (on a user's "favorites" page, for example) and get the associated blog objects via {{ favorite.content_object.title }}. What I want now, and can't figure out, is what I need to do to the blog model to allow me to have some tether to the favorite (so when it is displayed in a list it can be highlighted, for example). Here is the blog model: blog/models.py from django.db import models from django.db.models import permalink from django.template.defaultfilters import slugify from category.models import Category from section.models import Section from favorite.models import Favorite from django.contrib.auth.models import User from django.contrib.contenttypes.models import ContentType from django.contrib.contenttypes import generic class Blog(models.Model): title = models.CharField(max_length=200, unique=True) slug = models.SlugField(max_length=140, editable=False) author = models.ForeignKey(User) homepage = models.URLField() feed = models.URLField() description = models.TextField() page_views = models.IntegerField(null=True, blank=True, default=0 ) created_on = models.DateTimeField(auto_now_add = True) updated_on = models.DateTimeField(auto_now = True) def __unicode__(self): return self.title @models.permalink def get_absolute_url(self): return ('blog.views.show', [str(self.slug)]) def save(self, *args, **kwargs): if not self.slug: slug = slugify(self.title) duplicate_count = Blog.objects.filter(slug__startswith = slug).count() if duplicate_count: slug = slug + str(duplicate_count) self.slug = slug super(Blog, self).save(*args, **kwargs) class Entry(models.Model): blog = models.ForeignKey('Blog') title = models.CharField(max_length=200) slug = models.SlugField(max_length=140, editable=False) description = models.TextField() url = models.URLField(unique=True) image = models.URLField(blank=True, null=True) created_on = models.DateTimeField(auto_now_add = True) def __unicode__(self): return self.title def save(self, *args, **kwargs): if not self.slug: slug = slugify(self.title) duplicate_count = Entry.objects.filter(slug__startswith = slug).count() if duplicate_count: slug = slug + str(duplicate_count) self.slug = slug super(Entry, self).save(*args, **kwargs) class Meta: verbose_name = "Entry" verbose_name_plural = "Entries" Any guidance?

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