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  • In Django, how to define a "location path" in order to display it to the users?

    - by naw
    I want to put a "location path" in my pages showing where the user is. Supposing that the user is watching one product, it could be Index > Products > ProductName where each word is also a link to other pages. I was thinking on passing to the template a variable with the path like [(_('Index'), 'index_url_name'), (_('Products'), 'products_list_url_name'), (_('ProductName'), 'product_url_name')] But then I wonder where and how would you define the hierarchy without repeating myself (DRY)? As far I know I have seen two options To define the hierarchy in the urlconf. It could be a good place since the URL hierarchy should be similar to the "location path", but I will end repeating fragments of the paths. To write a context processor that guesses the location path from the url and includes the variable in the context. But this would imply to maintain a independient hierarchy wich will need to be kept in sync with the urls everytime I modify them. Also, I'm not sure about how to handle the urls that require parameters. Do you have any tip or advice about this? Is there any canonical way to do this?

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  • Cant get the child dir in django hosting (alwaysdata.com) .

    - by zjm1126
    this is my file : mysite templates homepage.html accounts a.html login_view.html i can get the homepage.html and accounts\a.html on 127.0.0.1:8000 but in http://zjm1126.alwaysdata.net , i can only get the homepage.html ,and cant get the account\a.html , this is my code : return render_to_response('accounts/login_view.html') and the accounts/login_view.html is : {% include "accounts\a.html" %} what can i do , thanks ,

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  • What is the difference between a site and an app in Django?

    - by larf311
    I know a site can have many apps but all the examples I see have the site called "mysite". I figured the site would be the name of your site, like StackOverflow for example. Would you do that and then have apps like "authentication", "questions", and "search"? Or would you really just have a site called mysite with one app called StackOverflow?

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  • Django: returning a selection of fields from a model based on their values?

    - by AP257
    I am working with some data over which I have little control. I'd like to return ONLY the fields of my model that aren't certain 'uninteresting' values (e.g. '0', 'X' or '-'), and access them individually in the template. My model is like this: class Manors(models.Model): structidx = models.IntegerField(primary_key=True, verbose_name="ID") hills = models.CharField(max_length=100, null=True, blank=True, verbose_name="Number of fields") In my template, I return a QuerySet of Manors, and I'd like to output something like this if the hills field isn't uninteresting: {% for manor in manors %} {% if manor.hills %}<li>Hills blah blah: {{ manor.hills }}</li>{% endif %} {% endfor %} I want to avoid too much logic in the template. Ideally, the manor object would simply not return with the uninteresting fields attached, then I could just do {% if manor.hills %}. I tried writing a model method that returns a dictionary of the interesting values, like this: def get_field_dictionary(self): interesting_fields = {} for field in Manors._meta.fields: if field.value_to_string(self) != "N" and field.value_to_string(self) != "0" and field.value_to_string(self) != "-" and field.value_to_string(self) != "X": interesting_fields[field.name] = field.value_to_string(self) return interesting_fields But I don't know how to access individual values of the dictionary in the template: {% if manor.get_field_dictionary['hills'] %}<li>Hills blah blah: {{ manor.get_field_dictionary['hills'] }}</li>{% endif %} gives a TemplateSyntaxError. Is there a better way to do this?

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  • Creating a User Registration Page using MongoEngine

    - by Drew Watkins
    I am currently working an a webapp, using mongoengine and django, which will require users to create an account from a registration page. I know MongoEngine has an authentication backend, but does it also include a registration form, etc..., like django itself does? If not, are there any example projects which show how to implement this? The only open-source mongoengine project I've found is django-mumblr, but I can't find the examples I want in it. I'm not interested in alternative options, such as MongoKit or mango for handling authentication. I am just getting started with django and mongoDB, so please excuse my lack of knowledge. Thanks in advance for the help!

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  • How do I use multiple settings file in Django with multiple sites on one server?

    - by William Bing Hua
    I have an ec2 instance running Ubuntu 14.04 and I want to host two sites from it. On my first site I have two settings file, production_settings.py and settings.py (for local development). I import the local settings into the production settings and override any settings with the production settings file. Since my production settings file is not the default settings.py name, I have to create an environment variable DJANGO_SETTINGS_MODULE='site1.production_settings' However because of this whenever I try to start my second site it says No module named site1.production_settings I am assuming that this is due to me setting the environment variable. Another problem is that I won't be able to use different settings file for different sites. How do I start use two different settings file for two different websites?

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  • Django Problem - trying to access data entered into a form and feed it through a different page

    - by John Hoke
    OK, so let me give you an overview first. I have this site and in it there is a form section. When you access that section you can view or start a new project. Each project has 3-5 different forms. My problem is that I don't want viewers to have to go through all 3-5 pages to see the relevant information they need. Instead I want to give each project a main page where all the essential data entered into the forms is shown as non-editable data. I hope this makes sense. So I need to find a way to access all that data from the different forms for each project and to feed that data into the new page I'll be calling "Main". Each project will have a separate main page for itself. I'm pretty much clueless as to how I should do this, so any help at all would be appreciated. Thanks

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  • How can I update only certain fields in a Django model form?

    - by J. Frankenstein
    I have a model form that I use to update a model. class Turtle(models.Model): name = models.CharField(max_length=50, blank=False) description = models.TextField(blank=True) class TurtleForm(forms.ModelForm): class Meta: model = Turtle Sometimes I don't need to update the entire model, but only want to update one of the fields. So when I POST the form only has information for the description. When I do that the model never saves because it thinks that the name is being blanked out while my intent is that the name not change and just be used from the model. turtle_form = TurtleForm(request.POST, instance=object) if turtle_form.is_valid(): turtle_form.save() Is there any way to make this happen? Thanks!

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  • Installing OSQA on windows(Local system)

    - by Pankaj Khurana
    Hi I want to install osqa on windows local system for this i have downloaded bitnami-djangostack-1.1.1-2-windows-installer.exe which has in built django,python,mysql & apache. I have run a django example given on the django website. Its working fine. But i am confused how to install osqa. I have downloaded the source code available on osqa site and readed the installation instruction(requires django 1.1.1). But how to make it working? Please help me on this Thanks

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  • Is there a better way to format this Python/Django code as valid PEP8?

    - by Ryan Detzel
    I have code written both ways and I see flaws in both of them. Is there another way to write this or is one approach more "correct" than the other? def functionOne(subscriber): try: results = MyModelObject.objects.filter( project__id=1, status=MyModelObject.STATUS.accepted, subscriber=subscriber).values_list( 'project_id', flat=True).order_by('-created_on') except: pass def functionOne(subscriber): try: results = MyModelObject.objects.filter( project__id=1, status=MyModelObject.STATUS.accepted, subscriber=subscriber) results = results.values_list('project_id', flat=True) results = results.order_by('-created_on') except: pass

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  • Best practice: How to persist simple data without a database in django?

    - by Infinity
    I'm building a website that doesn't require a database because a REST API "is the database". (Except you don't want to be putting site-specific things in there, since the API is used by mostly mobile clients) However there's a few things that normally would be put in a database, for example the "jobs" page. You have master list view, and the detail views for each job, and it should be easy to add new job entries. (not necessarily via a CMS, but that would be awesome) e.g. example.com/careers/ and example.com/careers/77/ I could just hardcode this stuff in templates, but that's no DRY- you have to update the master template and the detail template every time. What do you guys think? Maybe a YAML file? Or any better ideas? Thx

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  • How can I traverse a reverse generic relation in a Django template?

    - by user569139
    I have the following class that I am using to bookmark items: class BookmarkedItem(models.Model): is_bookmarked = models.BooleanField(default=False) user = models.ForeignKey(User) content_type = models.ForeignKey(ContentType) object_id = models.PositiveIntegerField() content_object = generic.GenericForeignKey() And I am defining a reverse generic relationship as follows: class Link(models.Model): url = models.URLField() bookmarks = generic.GenericRelation(BookmarkedItem) In one of my views I generate a queryset of all links and add this to a context: links = Link.objects.all() context = { 'links': links } return render_to_response('links.html', context) The problem I am having is how to traverse the generic relationship in my template. For each link I want to be able to check the is_bookmarked attribute and change the add/remove bookmark button according to whether the user already has it bookmarked or not. Is this possible to do in the template? Or do I have to do some additional filtering in the view and pass another queryset?

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  • [Django] How do I filter the choices in a ModelForm that has a CharField with the choices attribute

    - by nubela
    I understand I am able to filter queryset of Foreignkey or Many2ManyFields, however, how do I do that for a simple CharField that is a Select Widget (Select Tag). For example: PRODUCT_STATUS = ( ("unapproved", "Unapproved"), ("approved", "Listed"), #("Backorder","Backorder"), #("oos","Out of Stock"), #("preorder","Preorder"), ("userdisabled", "User Disabled"), ("disapproved", "Disapproved by admin"), ) and the Field: o_status = models.CharField(max_length=100, choices=PRODUCT_STATUS, verbose_name="Product Status", default="approved") Suppose I wish to limit it to just "approved" and "userdisabled" instead showing the full array (which is what I want to show in the admin), how do I do it? Thanks!

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