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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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  • 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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  • 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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  • Django exclude(**kwargs) help

    - by shawnjan
    Hey guys/gals! I had a question for you, something that I can't seem to find the solution for... Basically, I have a model called Environment, and I am passing all of them to a view, and there are particular environments that I would like to exclude. Now, I know there is a exclude function on a queryset, but I can't seem to figure out how to use it for multiple options... For example, I tried this but it didn't work: kwargs = {"name": "env1", "name": "env2"} envs = Environment.objects.exclude( kwards ) But the only thing that it will exclude is the last "name" value in the list of kwargs. I understand why it does that now, but I still can't seem to exclude multiple objects with one command. Any help is much appreciated! Shawn

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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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  • Caching query results in django

    - by Marcio Cruz
    I'm trying to find a way to cache the results of a query that won't change with frequency. For example, categories of products from an e-commerce (cellphones, TV, etc). I'm thinking of using the template fragment caching, but in this fragment, I will iterate over a list of these categories. This list is avaliable in any part of the site, so it's in my base.html file. Do I have always to send the list of categories when rendering the templates? Or is there a more dynamic way to do this, making the list always available in the template?

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  • Show choosen option in a notification Feed, Django

    - by apoo
    Hey I have a model where : LIST_OPTIONS = ( ('cheap','cheap'), ('expensive','expensive'), ('normal', 'normal'), ) then I have assigned the LIST_OPTIONS to nature variable. nature = models.CharField(max_length=15, choices=LIST_OPTIONS, null=False, blank=False). then I save it: if self.pk: new=False else: new=True super(Listing, self).save(force_insert, force_update) if new and notification: notification.send(User.objects.all().exclude(id=self.owner.id), "listing_new", {'listing':self, }, ) then in my management.py: def create_notice_types(app, created_models,verbosity, **kwargs): notification.create_notice_type("listing_new", _("New Listing"), _("someone has posted a new listing"), default=2) and now in my notice.html I want to show to users different sentences based on the options that they have choose so something like this: LINK href="{{ listing.owner.get_absolute_url }} {{listing.owner}} {% ifequal listing.nature "For Sale" %} created a {{ listing.nature }} listing, <a href="{{ listing.get_absolute_url }}">{{listing.title}}</a>. {% ifequals listing.equal "Give Away"%} is {{ listing.nature }} , LINK href="{{ listing.get_absolute_url }}" {{listing.title}}. {% ifequal listing.equal "Looking For"%} is {{ listing.nature }} , LINK href="{{ listing.get_absolute_url }}" {{listing.title}} {% endifequal %} {% endifequal %} {% endifequal %} Could you please help me out with this. Thank you

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  • Django ManyToMany join query

    - by Hanpan
    I'm sure this is really simple, but I can't for the life of me find any documentation explaining how to do this. How do I get the results of a ManyToMany field inside a join as opposed to doing this: {% for tag in article.tags.all %} Which results in an extra query? What I'd like to do is fetch all related tags when I retrieve the initial article, so I could then do something like: {% for tag in article.tags %} Without the .all and the extra query. Thanks!

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  • django ManyToMany through help

    - by dotty
    Hay I've got a question about relationships. I want to Users to have Friendships. So a User can be a friend with another User. I'm assuming i'll need to use the ManyToManyField, through a Friendship table. But i cannot get it to work. Any ideas? Here are my models. class User(models.Model): username = models.CharField(max_length=999) password = models.CharField(max_length=999) created_on = models.DateField(auto_now = False, auto_now_add = True) updated_on = models.DateField(auto_now = True, auto_now_add = False) friends = models.ManyToManyField('User', through='Friendship') class Friendship(models.Model): user = models.ForeignKey('User') friend = models.ForeignKey('User') Thanks

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  • Django inlineformset validation and delete

    - by Andrew Gee
    Hi, Can someone tell me if a form in an inlineformset should go through validation if the DELETE field is checked. I have a form that uses an inlineformset and when I check the DELETE box it fails because the required fields are blank. If I put data in the fields it will pass validation and then be deleted. Is that how it is supposed to work, I would have thought that if it is marked for delete it would bypass the validation for that form. Regards Andrew Follow up - but I would still appreciate some others opinions/help What I have figured out is that for validation to work the a formset form must either be empty or complete(valid) otherwise it will have errors when it is created and will not be deleted. As I have a couple of hidden fields in my formset forms and they are pre-populated when the page loads via javascript the form fails validation on the other required fields which might still be blank. The way I have gotten around this by adding in a check in the add_fields that tests if the DELETE input is True and if it is it makes all fields on the form not required, which means it passes validation and will then delete. def add_fields(self, form, index) #add other fields that are required.... deleteValue = form.fields['DELETE'].widget.value_from datadict(form.data, form.files, form.add_prefix('DELETE')) if bool(deleteValue) or deleteValue == '': for name, field in form.fields.items(): form.fields[name].required= False This seems to be an odd way to do things but I cannot figure out another way. Is there a simpler way that I am missing? I have also noticed that when I add the new form to my page and check the Delete box, there is no value passed back via the request, however an existing form (one loaded from the database) has a value of on when the Delete box is checked. If the box is not checked then the input is not in the request at all. Thanks Andrew

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  • django error:The model Tribe is already registered

    - by zjm1126
    when i python manage.py syncdb,it show this: The following content types are stale and need to be deleted: maps | tribe Any objects related to these content types by a foreign key will also be deleted. Are you sure you want to delete these content types? If you're unsure, answer 'no'. Type 'yes' to continue, or 'no' to cancel: no when i put 'no' ,and then python manage runserver: AlreadyRegistered at / The model Tribe is already registered what should i do ? thanks

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  • onmouseover with django / imagekit

    - by Michael Moreno
    I'm using Imagekit. View.py includes: def pics(request): p = Photo.objects.all() return render_to_response('Shots.html', {'p': p}) The following simple code in the template will generate associated images: {% for p in p %} <img src = "{{ p.display.url }}"> <img src = "{{ p.thumbnail_image.url }}"> {% endfor %} I'm attempting to generate a series of thumbnails {{ p.thumbnail_image.url }} which, when mouseover'd, will generate the slightly larger version of the image, {{ p.display.url }} via Javascript. The following code in the template attempts to do so: <html> <head> <HEAD> <script language="Javascript"> { image1 = new Image image2 = new Image image1.src = {{ p.thumbnail_image.url }} image2.src = {{ p.display.url }} </script> </head> <body> {% for p in p %} <a href="" onMouseOver="document.rollover.src= image2.src onMouseOut="document.rollover.src= image1.src"> <img src="{{ p.thumbnail_image.url }}" border=0 name="rollover"></a> {% endfor %} </body> </html> This will display the series of thumbnails, but the larger image will not display when mouseover'd. I believe it has to do with how I'm specifying the variable {{ p.display.url }}.

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  • django: search forms and redirect

    - by gruszczy
    After processing form from POST I should redirect, to prevent user from hitting back. However, I am using form to determine search query on a database, so I need to either pass params to the redirected site or the result of a search. Or maybe there is some other good practice, how to solve this problem? Maybe in this situation I am allowed not to redirect (nothing happens, if user performs search again).

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  • Timezones and the DateTimeField - Django

    - by RadiantHex
    Hi folks, I'm trying to implement a "time ago" feature, for the displaying of items on a site. As I'm caching the pages I wish to use javascript in order to render the "time ago". Javascript knows local time and problably the Timezone of the local machine so I could play with that, but that would require to hard code the server's timezone. Therefore I'm trying to figure out a simple way to pass a ISO 8601 timestamp, in GMT time. Is there any simple and straight forward way for doing this? Help would be much appreciated! =)

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  • Django: Odd mark_safe behaviour?

    - by Mark
    I wrote this little function for writing out HTML tags: def html_tag(tag, content=None, close=True, attrs={}): lst = ['<',tag] for key, val in attrs.iteritems(): lst.append(' %s="%s"' % (key, escape_html(val))) if close: if content is None: lst.append(' />') else: lst.extend(['>', content, '</', tag, '>']) else: lst.append('>') return mark_safe(''.join(lst)) Which worked great, but then I read this article on efficient string concatenation (I know it doesn't really matter for this, but I wanted consistency) and decided to update my script: def html_tag(tag, body=None, close=True, attrs={}): s = StringIO() s.write('<%s'%tag) for key, val in attrs.iteritems(): s.write(' %s="%s"' % (key, escape_html(val))) if close: if body is None: s.write(' />') else: s.write('>%s</%s>' % (body, tag)) else: s.write('>') return mark_safe(s.getvalue()) But now my HTML get escaped when I try to render it from my template. Everything else is exactly the same. It works properly if I replace the last line with return mark_safe(unicode(s.getvalue())). I checked the return type of s.getvalue(). It should be a str, just like the first function, so why is this failing?? Also fails with SafeString(s.getvalue()) but succeeds with SafeUnicode(s.getvalue()). I'd also like to point out that I used return mark_safe(s.getvalue()) in a different function with no odd behavior. The "call stack" looks like this: class Input(Widget): def render(self): return html_tag('input', attrs={'type':self.itype, 'id':self.id, 'name':self.name, 'value':self.value, 'class':self.itype}) class Field: def __unicode__(self): return mark_safe(self.widget.render()) And then {{myfield}} is in the template. So it does get mark_safed'd twice, which I thought might have been the problem, but I tried removing that too..... I really have no idea what's causing this, but it's not too hard to work around, so I guess I won't fret about it.

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  • Django model: Reference foreign key table in __unicode__ function for admin

    - by pa
    Example models: class Parent(models.Model): name = models.CharField() def __unicode__(self): return self.name class Child(models.Model): parent = models.ForeignKey(Parent) def __unicode__(self): return self.parent.name # Would reference name above I'm wanting the Child.unicode to refer to Parent.name, mostly for the admin section so I don't end up with "Child object" or similar, I'd prefer to display it more like "Child of ". Is this possible? Most of what I've tried hasn't worked unfortunately.

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