Search Results

Search found 9098 results on 364 pages for 'django admin'.

Page 16/364 | < Previous Page | 12 13 14 15 16 17 18 19 20 21 22 23  | Next Page >

  • 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.

    Read the article

  • How can I run .aggregate() on a field introduced using .extra(select={...}) in a Django Query?

    - by Jake
    I'm trying to get the count of the number of times a player played each week like this: player.game_objects.extra(select={'week': 'WEEK(`games_game`.`date`)'}).aggregate(count=Count('week')) But Django complains that FieldError: Cannot resolve keyword 'week' into field. Choices are: <lists model fields> I can do it in raw SQL like this SELECT WEEK(date) as week, COUNT(WEEK(date)) as count FROM games_game WHERE player_id = 3 GROUP BY week Is there a good way to do this without executing raw SQL in Django?

    Read the article

  • wierd FileField url in admin site

    - by panchicore
    My model class TheFile(models.Model): document = models.FileField(upload_to="archivos") The wierd HTML admin link: <a hacking_google_maps_and_google_earth.pdf="" archivos="" media="" localhost:8000="" http:="" href="" target="_blank">archivos/Hacking_Google_Maps_And_Google_Earth.pdf</a> If I firebug-edit the href="" it works :S

    Read the article

  • weird FileField url in admin site

    - by panchicore
    My model class TheFile(models.Model): document = models.FileField(upload_to="archivos") The weird HTML admin link: <a hacking_google_maps_and_google_earth.pdf="" archivos="" media="" localhost:8000="" http:="" href="" target="_blank">archivos/Hacking_Google_Maps_And_Google_Earth.pdf</a> If I firebug-edit the href="" it works :S

    Read the article

  • how to customize django admin for clickable list_editable

    - by FurtiveFelon
    Hi all, Currently, i have a class MyAdmin(admin.ModelAdmin), and i have a field in there called name, which belongs to both list_editable and list_display. The current behavior is such that all fields that is in list_editable displays a form field. However, i would like to change that only when people click on the field would it turn into a editable form field. Can anyone point me in the right direction on how to do that (which template to edit etc.). Thank you very much! Jason

    Read the article

  • SelectDateWidget in Django Admin?

    - by Maria
    Can i change default AdminDateWidget to SelectDateWidget in my models? How can i do this? I try: class RespondentAdmin(admin.ModelAdmin): formfield_overrides = { models.DateField: {'widget': SelectDateWidget}, } but it doesn't work

    Read the article

  • django admin site - filtering available objects for user

    - by JPG
    I have models that belong to some 'group' (Company class). I want to add users, who will also belong to a one group and should be able to edit/manage/add objects with membership in associated group. something like: class Company() class Something() company = ForeignKey(Company) user Microsoft_admin company = ForeignKey(Company) and this user should only see and edit objects belonging to associated Company in the Admin Interface. How to acomplish that?

    Read the article

  • Django admin default filter

    - by h3
    I know I already managed to do this but can't remember how nor I can't find any documentation about this.. How can apply a filter by default on a object list view in the admin ? I have an app which list quotes and those quotes have a status (ex: accepted, rejected, on hold ..). I want the filter set on status='accepted' by default that is..

    Read the article

  • Django Custom Admin

    - by Harry
    Hello there Currently I have an app with a save method in its models.py . So I would like to change this so the models does not do the save() method, but rather have a separate view that will do this in die admin site. Can you please direct me in the correct direction? Thank you

    Read the article

  • How can I download django-1.2 and use it across multiple sites when the system default is 1.1?

    - by meder
    I'm on Debian Lenny and the latest backports django is 1.1.1 final. I don't want to use sid so I probably have to download django. I have my sites located at: /www/ and I plan on using mod_wsgi with Apache2 as a reverse proxy from nginx. Now that I downloaded pip and virtualenv through pip, can someone explain how I could get my /www/ sites which are yet to be made to all use django-1.2? Question 1.1: Where do you suggest I download django-1.2? I know you can store it anywhere but where would you store it? Question 1.2: After installing it how do you actually tie that django-1.2 instead of the system default django 1.2 to the reverse proxied Apache conf? I would prefer it if answers were more specific than vague and have examples of setups.

    Read the article

  • djnago-multilingual-ng / Django 1.1.1 incompatibility?

    - by omat
    I am getting "cannot import name connections" exception while trying to use django-multilingual-ng 0.1.20 with Django 1.1.1. The exception comes from the line 15 of query.py where it tries to: from django.db import connections, DEFAULT_DB_ALIAS Is it not compatible with Django 1.1.1? Does anybody tried this combination and have any suggestions? Thanks. -- oMat

    Read the article

  • Extending Django Flatpages to accept template tags

    - by Tristan
    I use django flatpages for a lot of content on our site, I'd like to extend it to accept django template tags in the content as well. I found this snippet but after much larking about I couldn't get it to work. Am I correct in assuming that you would need too "subclass" the django flatpages app to get this to work? Is this best way of doing it? I'm not quite sure how to structure it, as I don't really want to directly modify the django distribution.

    Read the article

  • Django ModelFormSet with Google app engine

    - by Eric
    I'm using Django with google app engine. I'm using the google furnished django app engine helper project. I'm attempting to create a Django modelformset like this: #MyModel inherits from BaseModel MyFormSet = modelformset_factory(models.MyModel) However, it's failing with this error: 'ModelOptions' object has no attribute 'fields' Apparently modelformset_factory() is expecting MyModel to implement a 'fields' accessor. Anybody successfully used a modelformset with GAE datastore? Or is this a fundamental incompatibility between Django and GAE?

    Read the article

  • django-registration password reset custom template not loading

    - by ip.
    I'm using django-registration for registering users, however when I want to use my own template for password reset I get the admin template and not the template I created. My template is in myapp/templates/registration/password_reset_form.html and my template loaders are properly set: TEMPLATE_LOADERS = ( 'django.template.loaders.filesystem.Loader', 'django.template.loaders.app_directories.Loader', ) What could I be missing? I'm using Django 1.4

    Read the article

  • Learning django by example source code (not examples)

    - by Bryce
    I'm seeking a nice complete open source django application to study and learn best practices from, or even use as a template. The tutorials only go so far, and django is super flexible which can lead one to paining themselves into a corner. Ideally such a template / example would: Ignore django admin, and implement full CRUD outside the admin. Be built like a large application in terms of best practices and patterns. Have a unit test Use at least one package (e.g. twitter integration or threaded comments) Implement some AJAX or Comet See also: Learning Django by example

    Read the article

  • Visual web page designer for Django?

    - by Robert Oschler
    I'm just starting my Django learning so pardon me if any part of this question is off-base. I have done a lot of web searching for information on the equivalent of a visual web page designer for Django and I don't seem to be getting very far. I have experience with Delphi (Object Pascal), C, C++, Python, PHP, Java, and Javascript and have created and maintained several web sites that included MySQL database dependent content. For the longest time I've been using one of the standard WYSIWIG designers to design the actual web pages, with any needed back end programming done via Forms or AJAX calls that call server side PHP scripts. I have grown tired of the quirks, bugs, and other annoyances associated with the program. Also, I find myself hungry for the functionality and reliability a good MVC based framework would provide me so I could really express myself with custom code easily. So I am turning to Django/Python. However, I'm still a junkie for a good WYSIWIG designer for the layout of web pages. I know there are some out there that thrive on opening up a text editor, possibly with some code editor tools to assist, and pounding out pages. But I do adore a drag and drop editor for simple page layout, especially for things like embedded images, tables, buttons, etc. I found a few open source projects on GitHub but they seem to be focused on HTML web forms, not a generic web page editor. So can I get what I want here? The supreme goal would be to find not only a web page editor that creates Django compatible web pages, but if I dare say it, have a design editor that could add Python code stubs to various page elements in the style of the Delph/VCL or VB design editors. Note, I also have the Wing IDE Professional IDE, version 2.0. As a side note, if you know of any really cool, fun, or time-saving Python libraries that are designed for easy integration into Django please tell me about them. -- roschler

    Read the article

  • Does this syntax for specifying Django conditional form display align with python/django convention?

    - by andy
    I asked a similar question on Stackoverflow and was told it was better asked here. So I'll ask it slightly rephrased. I am working on a Django project, part of which will become a distributable plugin that allows the python/django developer to specify conditional form field display logic in the form class or model class. I am trying to decide how the developer must specify that logic. Here's an example: class MyModel(models.Model): #these are some django model fields which will be used in a form yes_or_no = models.SomeField...choices are yes or no... why = models.SomeField...text, but only relevant if yes_or_no == yes... elaborate_even_more = models.SomeField...more text, just here so we can have multiple conditions #here i am inventing some syntax...i am looking for suggestions!! #this is one possibility why.show_if = ('yes_or_no','==','yes') elaborate_even_more.show_if = (('yes_or_no','==','yes'),('why','is not','None')) #help me choose a syntax that is *easy*...and Pythonic and...Djangonic...and that makes your fingers happy to type! #another alternative... conditions = {'why': ('yes_or_no','==','yes'), 'elaborate_even_more': (('yes_or_no','==','yes'),('why','is not','None')) } #or another alternative... """Showe the field whiche hath the name *why* only under that circumstance in whiche the field whiche hath the name *yes_or_no* hath the value *yes*, in strictest equality.""" etc... Those conditions will be eventually passed via django templates to some javascript that will show or hide form fields accordingly. Which of those options (or please propose a better option) aligns better with conventions such that it will be easiest for the python/django developer to use? Also are there other considerations that should impact what syntax I choose?

    Read the article

  • Rebuilding website from Django 0.96 to Django 1.2

    - by Neytiri
    I've got a website done in Django 0.96 (done in 2007), and now we are thinking about rebuilding it (not just migrating) for Django 1.2 . Can anyone point me to the new (and worth the while) widgets, plugins and other stuff for Django 1.2 (released in april 2010). I've heard of "South" and of a widget for debugging (can't remember the name), but I'm a little lost here.

    Read the article

  • Why use Django on Google App Engine?

    - by Travis Bradshaw
    When researching Google App Engine (GAE), it's clear that using Django is wildly popular for developing in Python on GAE. I've been scouring the web to find information on the costs and benefits of using Django, to find out why it's so popular. While I've been able to find a wide variety of sources on how to run Django on GAE and the various methods of doing so, I haven't found any comparative analysis on why Django is preferable to using the webapp framework provided by Google. To be clear, it's immediately apparent why using Django on GAE is useful for developers with an existing skillset in Django (a majority of Python web developers, no doubt) or existing code in Django (where using GAE is more of a porting exercise). My team, however, is evaluating GAE for use on an all-new project and our existing experience is with TurboGears, not Django. It's been quite difficult to determine why Django is beneficial to a development team when the BigTable libraries have replaced Django's ORM, sessions and authentication are necessarily changed, and Django's templating (if desirable) is available without using the entire Django stack. Finally, it's clear that using Django does have the advantage of providing an "exit strategy" if we later wanted to move away from GAE and need a platform to target for the exodus. I'd be extremely appreciative for help in pointing out why using Django is better than using webapp on GAE. I'm also completely inexperienced with Django, so elaboration on smaller features and/or conveniences that work on GAE are also valuable to me. Thanks in advance for your time!

    Read the article

  • Django vs GAE + Django vs GAE + other framework

    - by Ilian Iliev
    I`m looking for opinion which one is better for building web applications(web sites). I have some experience with Django, and some with Google App Engine and App-Engine-Patch for Django. And it seems to me that only Django is working faster than the GAE implementation. Is there some other frameworks that simplify the developments process, providing forms creating, user management, url resolving etc. Thanks in advance, Ilian Iliev P.S. I am also interested in GAE and webapp framework case

    Read the article

  • Google App Engine + Form Validation

    - by Iwona
    Hi, I would like to do google app engine form validation but I dont know how to do it? I tried like this: from google.appengine.ext.db import djangoforms from django import newforms as forms class SurveyForm(forms.Form): occupations_choices = ( ('1', ""), ('2', "Undergraduate student"), ('3', "Postgraduate student (MSc)"), ('4', "Postgraduate student (PhD)"), ('5', "Lab assistant"), ('6', "Technician"), ('7', "Lecturer"), ('8', "Other" ) ) howreach_choices = ( ('1', ""), ('2', "Typed the URL directly"), ('3', "Site is bookmarked"), ('4', "A search engine"), ('5', "A link from another site"), ('6', "From a book"), ('7', "Other") ) boxes_choices = ( ("des", "Website Design"), ("svr", "Web Server Administration"), ("com", "Electronic Commerce"), ("mkt", "Web Marketing/Advertising"), ("edu", "Web-Related Education") ) name = forms.CharField(label='Name', max_length=100, required=True) email = forms.EmailField(label='Your Email Address:') occupations = forms.ChoiceField(choices=occupations_choices, label='What is your occupation?') howreach = forms.ChoiceField(choices=howreach_choices, label='How did you reach this site?') # radio buttons 1-5 rating = forms.ChoiceField(choices=range(1,6), label='What is your occupation?', widget=forms.RadioSelect) boxes = forms.ChoiceField(choices=boxes_choices, label='Are you involved in any of the following? (check all that apply):', widget=forms.CheckboxInput) comment = forms.CharField(widget=forms.Textarea, required=False) And I wanted to display it like this: template_values = { 'url' : url, 'url_linktext' : url_linktext, 'userName' : userName, 'item1' : SurveyForm() } And I have this error message: Traceback (most recent call last): File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp_init_.py", line 515, in call handler.get(*groups) File "C:\Program Files\Google\google_appengine\demos\b00213576\main.py", line 144, in get self.response.out.write(template.render(path, template_values)) File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 143, in render return t.render(Context(template_dict)) File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 183, in wrap_render return orig_render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 168, in render return self.nodelist.render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 705, in render bits.append(self.render_node(node, context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 718, in render_node return(node.render(context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template\defaulttags.py", line 209, in render return self.nodelist_true.render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 705, in render bits.append(self.render_node(node, context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 718, in render_node return(node.render(context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 768, in render return self.encode_output(output) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 757, in encode_output return str(output) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\util.py", line 26, in str return self.unicode().encode(settings.DEFAULT_CHARSET) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 73, in unicode return self.as_table() File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 144, in as_table return self._html_output(u'%(label)s%(errors)s%(field)s%(help_text)s', u'%s', '', u'%s', False) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 129, in _html_output output.append(normal_row % {'errors': bf_errors, 'label': label, 'field': unicode(bf), 'help_text': help_text}) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 232, in unicode value = value.str() File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\util.py", line 26, in str return self.unicode().encode(settings.DEFAULT_CHARSET) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 246, in unicode return u'\n%s\n' % u'\n'.join([u'%s' % w for w in self]) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 238, in iter yield RadioInput(self.name, self.value, self.attrs.copy(), choice, i) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 212, in init self.choice_value = smart_unicode(choice[0]) TypeError: 'int' object is unsubscriptable Do You have any idea how I can do this validation in different case? I have tried to do it using this kind of: class ItemUserAnswer(djangoforms.ModelForm): class Meta: model = UserAnswer But I dont know how to add extra labels to this form and it is displayed in one line. Do You have any suggestions? Thanks a lot as it making me crazy why it is still not working:/

    Read the article

  • How to limit choice field options based on another choice field in django admin

    - by umnik700
    I have the following models: class Category(models.Model): name = models.CharField(max_length=40) class Item(models.Model): name = models.CharField(max_length=40) category = models.ForeignKey(Category) class Demo(models.Model): name = models.CharField(max_length=40) category = models.ForeignKey(Category) item = models.ForeignKey(Item) In the admin interface when creating a new Demo, after user picks category from the dropdown, I would like to limit the number of choices in the "items" drop-down. If user selects another category then the item choices should update accordingly. I would like to limit item choices right on the client, before it even hits the form validation on the server. This is for usability, because the list of items could be 1000+ being able to narrow it down by category would help to make it more manageable. Is there a "django-way" of doing it or is custom JavaScript the only option here?

    Read the article

  • how to change display text in django admin foreignkey dropdown

    - by FurtiveFelon
    Hi all, I have a task list, with ability to assign users. So i have foreignkey to User model in the database. However, the default display is username in the dropdown menu, i would like to display full name (first last) instead of the username. If the foreignkey is pointing to one of my own classes, i can just change the str function in the model, but User is a django authentication model, so i can't easily change it directly right? Anyone have any idea how to accomplish this? Thanks a lot!

    Read the article

  • Django admin page dropdowns

    - by zen
    I am building a high school team application using Django. Here is my working models file: class Directory(models.Model): school = models.CharField(max_length=60) website = models.URLField() district = models.SmallIntegerField() conference = models.ForeignKey(Conference) class Conference(models.Model): conference_name = models.CharField(max_length=50) url = models.URLField() class Meta: ordering = ['conference_name'] When I open my admin pages and go to edit a school's conference the drop down looks like this: <select> <option value="1">Conference Object</option> <option value="2">Conference Object</option> <select> How do I replace "Conference Object" with the conference_name?

    Read the article

  • Django admin return to page after save

    - by Thordin9
    Hi all, I have 3 pages of items listed in my django application admin. After i edit one of them (lets say it is in page 2) and save my changes, i return to page 1 of my listing. How can i make it so i return to the page the item is in? I looked into some similar questions here at stackoverflow and i believe that i need to use javascript to send a httpresponse with the location header. But how i can determine the page the item is in? any help is appreciated

    Read the article

< Previous Page | 12 13 14 15 16 17 18 19 20 21 22 23  | Next Page >