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  • Adding interactions to admin pages generated by the admin generator

    - by Stick it to THE MAN
    I am using Symfony 1.2.9 (with Propel ORM) to create a website. I have started using the admin generator to implement the admin functionality. I have come accross a slight 'problem' however. My models are related (e.g. one table may have several 1:N relations and N:N relations). I have not found a way to address this satisfactorily yet. As a tactical solution (for list views), I have decided to simply show the parent object, and then add interactions to show the related objects. I'll use a Blog model to illustrate this. Here are the relationships for a blog model: N:M relationship with Blogroll (models a blog roll) 1:N relationship with Blogpost (models a post submitted to a blog) I had originally intended on displaying the (paged) blogpost list for a blog,, when it was selected, using AJAX, but I am struggling enough with the admin generator as it is, so I have shelved that idea - unless someone is kind enough to shed some light on how to do this. Instead, what I am now doing (as a tactical/interim soln), is I have added interactions to the list view which allow a user to: View a list of the blog roll for the blog on that row View a list of the posts for the blog on that row Add a post for the blog on tha row In all of the above, I have written actions that will basically forward the request to the approriate action (admin generated). However, I need to pass some parameters (like the blog id etc), so that the correct blog roll or blog post list etc is returned. I am sure there is a better way of doing what I want to do, but in case there isn't here are my questions: How may I obtain the object that relates to a specific row (of the clicked link) in the list view (e.g. the blog object in this example) Once I have the object, I may choose to extract various fields: id etc. How can I pass these arguments to the admin generated action ? Regarding the second question, my guess is that this may be the way to do it (I may be wrong) public function executeMyAddedBlogRollInteractionLink(sfWebRequest $request) { // get the object *somehow* (I'm guessing this may work) $object = $this->getRoute()->getObject(); // retrieve the required parameters from the object, and build a query string $query_str=$object->getId(); //forward the request to the generated code (action to display blogroll list in this case) $this->forward('backendmodulename',"getblogrolllistaction?params=$query_string"); } This feels like a bit of a hack, but I'm not sure how else to go about it. I'm also not to keen on sending params (which may include user_id etc via a GET, even a POST is not that much safer, since it is fairly sraightforward to see what requests a browser is making). if there is a better way than what I suggest above to implement this kind of administration that is required for objects with 1 or more M:N relationships, I will be very glad to hear the "recommended" way of going about it. I remember reading about marking certain actions as internal. i.e. callable from only within the app. I wonder if that would be useful in this instance?

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  • Changing User ModelAdmin for Django admin

    - by Leon
    How do you override the admin model for Users? I thought this would work but it doesn't? class UserAdmin(admin.ModelAdmin): list_display = ('email', 'first_name', 'last_name') list_filter = ('is_staff', 'is_superuser') admin.site.register(User, UserAdmin) I'm not looking to override the template, just change the displayed fields & ordering. Solutions please?

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  • How do I secure all the admin actions in all controllers in cakePHP

    - by Gaurav Sharma
    Hello Everyone, I am developing an application using cakePHP v 1.3 on windows (XAMPP). Most of the controllers are baked with the admin routing enabled. I want to secure the admin actions of every controller with a login page. How can I do this without repeating much ? One solution to the problem is that "I check for login information in the admin_index action of every controller" and then show the login screen accordingly. Is there any better way of doing this ? The detault URL to admin (http://localhost/app/admin) is pointing to the index_admin action of users controller (created a new route for this in routes.php file) Thanks

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  • Hide fields in Django admin

    - by jwesonga
    I'm tying to hide my slug fields in the admin by setting editable=False but every time I do that I get the following error: KeyError at /admin/website/program/6/ Key 'slug' not found in Form Request Method: GET Request URL: http://localhost:8000/admin/website/program/6/ Exception Type: KeyError Exception Value: Key 'slug' not found in Form Exception Location: c:\Python26\lib\site-packages\django\forms\forms.py in __getitem__, line 105 Python Executable: c:\Python26\python.exe Python Version: 2.6.4 Any idea why this is happening

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  • Dropdown sorting in django-admin

    - by Andrey
    I'd like to know how can I sort values in the Django admin dropdowns. For example, I have a model called Article with a foreign key pointing to the Users model, smth like: class Article(models.Model): title = models.CharField(_('Title'), max_length=200) slug = models.SlugField(_('Slug'), unique_for_date='publish') author = models.ForeignKey(User) body = models.TextField(_('Body')) status = models.IntegerField(_('Status')) categories = models.ManyToManyField(Category, blank=True) publish = models.DateTimeField(_('Publish date')) I edit this model in django admin: class ArticleAdmin(admin.ModelAdmin): list_display = ('title', 'publish', 'status') list_filter = ('publish', 'categories', 'status') search_fields = ('title', 'body') prepopulated_fields = {'slug': ('title',)} admin.site.register(Article, ArticleAdmin) and of course it makes the nice user select dropdown for me, but it's not sorted and it takes a lot of time to find a user by username.

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  • Django admin's filter_horizontal (& filter_vertical) not working

    - by negus
    I'm trying to use ModelAdmin.filter_horizontal and ModelAdmin.filter_vertical for ManyToMany field instead of select multiple box but all I get is: My model: class Title(models.Model): #... production_companies = models.ManyToManyField(Company, verbose_name="????????-?????????????") #... My admin: class TitleAdmin(admin.ModelAdmin): prepopulated_fields = {"slug": ("original_name",)} filter_horizontal = ("production_companies",) radio_fields = {"state": admin.HORIZONTAL} #... The javascripts are loading OK, I really don't get what happens. Django 1.1.1 stable.

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  • where is everything in django admin?

    - by FurtiveFelon
    Hi all, I would like to figure out where everything is in django admin. Since i am currently trying to modify the behavior rather heavily right now, so perhaps a reference would be helpful. For example, where is ModelAdmin located, i cannot find it anywhere in C:\Python26\Lib\site-packages\django\contrib\admin. I need that because i would like to look at how it is implemented so that i can override with confidence. I need to do that in part because of this page: http://docs.djangoproject.com/en/dev/ref/contrib/admin/#modeladmin-methods, for example, i would like to override ModelAdmin.add_view, but i can't find the original source for that. As well as i would like to see the url routing file for admin, so i can easily figure out which url corresponding to which template etc. Thanks a lot for any pointers!

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  • Admin auto generation

    - by Me-and-Coding
    Hi, I create custom CMS sites and then go ahead for the backend/admin side. Is there any tool out there to automatically create admin side of my sites, for example, based on table relationships or whatever customization we may put in place. PHPMaker seems to claim something like what i ask for but i have not used it. Any tool out there to auto-create admin side or PHPMaker is up to the point? Thanks

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  • Building Admin Areas in Rails - General Questions

    - by Carb
    What is the typical format/structure for creating an administrative area in a Rails application? Specifically I am stumped in the vicinity of these topics: How do you deal with situations where a model's resources are available to both the public and the Admin? i.e. A User model where anyone can create users, login, etc but only the admin can view users, delete/update them, etc. What is the proper convention for routing? How does one structure controllers? Are duplicate controllers considered OK? i.e. An admin version and the non-admin version? Thank you!

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  • Django admin - remove field if editing an object

    - by John McCollum
    I have a model which is accessible through the Django admin area, something like the following: # model class Foo(models.Model): field_a = models.CharField(max_length=100) field_b = models.CharField(max_length=100) # admin.py class FooAdmin(admin.ModelAdmin): pass Let's say that I want to show field_a and field_b if the user is adding an object, but only field_a if the user is editing an object. Is there a simple way to do this, perhaps using the fields attribute? If if comes to it, I could hack a JavaScript solution, but it doesn't feel right to do that at all!

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  • Django admin panel doesn't work after modify default user model.

    - by damienix
    I was trying to extend user profile. I founded a few solutions, but the most recommended was to create new user class containing foreign key to original django.contrib.auth.models.User class. I did it with this so i have in models.py: class UserProfile(models.Model): user = models.ForeignKey(User, unique=True) website_url = models.URLField(verify_exists=False) and in my admin.py from django.contrib import admin from someapp.models import * from django.contrib.auth.admin import UserAdmin # Define an inline admin descriptor for UserProfile model class UserProfileInline(admin.TabularInline): model = UserProfile fk_name = 'user' max_num = 1 # Define a new UserAdmin class class MyUserAdmin(UserAdmin): inlines = [UserProfileInline, ] # Re-register UserAdmin admin.site.unregister(User) admin.site.register(User, MyUserAdmin) And now when I'm trying to create/edit user in admin panel i have an error: "Unknown column 'content_userprofile.id' in 'field list'" where content is my appname. I was trying to add line AUTH_PROFILE_MODULE = 'content.UserProfile' to my settings.py but with no effect. How to tell panel admin to know how to correctly display fields in user form?

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  • Custom Django admin URL + changelist view for custom list filter by Tags

    - by Botondus
    In django admin I wanted to set up a custom filter by tags (tags are introduced with django-tagging) I've made the ModelAdmin for this and it used to work fine, by appending custom urlconf and modifying the changelist view. It should work with URLs like: http://127.0.0.1:8000/admin/reviews/review/only-tagged-vista/ But now I get 'invalid literal for int() with base 10: 'only-tagged-vista', error which means it keeps matching the review edit page instead of the custom filter page, and I cannot figure out why since it used to work and I can't find what change might have affected this. Any help appreciated. Relevant code: class ReviewAdmin(VersionAdmin): def changelist_view(self, request, extra_context=None, **kwargs): from django.contrib.admin.views.main import ChangeList cl = ChangeList(request, self.model, list(self.list_display), self.list_display_links, self.list_filter, self.date_hierarchy, self.search_fields, self.list_select_related, self.list_per_page, self.list_editable, self) cl.formset = None if extra_context is None: extra_context = {} if kwargs.get('only_tagged'): tag = kwargs.get('tag') cl.result_list = cl.result_list.filter(tags__icontains=tag) extra_context['extra_filter'] = "Only tagged %s" % tag extra_context['cl'] = cl return super(ReviewAdmin, self).changelist_view(request, extra_context=extra_context) def get_urls(self): from django.conf.urls.defaults import patterns, url urls = super(ReviewAdmin, self).get_urls() def wrap(view): def wrapper(*args, **kwargs): return self.admin_site.admin_view(view)(*args, **kwargs) return update_wrapper(wrapper, view) info = self.model._meta.app_label, self.model._meta.module_name my_urls = patterns('', # make edit work from tagged filter list view # redirect to normal edit view url(r'^only-tagged-\w+/(?P<id>.+)/$', redirect_to, {'url': "/admin/"+self.model._meta.app_label+"/"+self.model._meta.module_name+"/%(id)s"} ), # tagged filter list view url(r'^only-tagged-(P<tag>\w+)/$', self.admin_site.admin_view(self.changelist_view), {'only_tagged':True}, name="changelist_view"), ) return my_urls + urls Edit: Original issue fixed. I now receive 'Cannot filter a query once a slice has been taken.' for line: cl.result_list = cl.result_list.filter(tags__icontains=tag) I'm not sure where this result list is sliced, before tag filter is applied. Edit2: It's because of the self.list_per_page in ChangeList declaration. However didn't find a proper solution yet. Temp fix: if kwargs.get('only_tagged'): list_per_page = 1000000 else: list_per_page = self.list_per_page cl = ChangeList(request, self.model, list(self.list_display), self.list_display_links, self.list_filter, self.date_hierarchy, self.search_fields, self.list_select_related, list_per_page, self.list_editable, self)

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  • Django - Override admin site's login form

    - by TrojanCentaur
    I'm currently trying to override the default form used in Django 1.4 when logging in to the admin site (my site uses an additional 'token' field required for users who opt in to Two Factor Authentication, and is mandatory for site staff). Django's default form does not support what I need. Currently, I've got a file in my templates/ directory called templates/admin/login.html, which seems to be correctly overriding the template used with the one I use throughout the rest of my site. The contents of the file are simply as below: # admin/login.html: {% extends "login.html" %} The actual login form is as below: # login.html: {% load url from future %}<!DOCTYPE html> <html> <head> <title>Please log in</title> </head> <body> <div id="loginform"> <form method="post" action="{% url 'id.views.auth' %}"> {% csrf_token %} <input type="hidden" name="next" value="{{ next }}" /> {{ form.username.label_tag }}<br/> {{ form.username }}<br/> {{ form.password.label_tag }}<br/> {{ form.password }}<br/> {{ form.token.label_tag }}<br/> {{ form.token }}<br/> <input type="submit" value="Log In" /> </form> </div> </body> </html> My issue is that the form provided works perfectly fine when accessed using my normal login URLs because I supply my own AuthenticationForm as the form to display, but through the Django Admin login route, Django likes to supply it's own form to this template and thus only the username and password fields render. Is there any way I can make this work, or is this something I am just better off 'hard coding' the HTML fields into the form for?

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  • Django internationalization for admin pages - translate model name and attributes

    - by geekQ
    Django's internationalization is very nice (gettext based, LocaleMiddleware), but what is the proper way to translate the model name and the attributes for admin pages? I did not find anything about this in the documentation: http://docs.djangoproject.com/en/dev/topics/i18n/internationalization/ http://www.djangobook.com/en/2.0/chapter19/ I would like to have "???????? ????? ??? ?????????" instead of "???????? order ??? ?????????". Note, the 'order' is not translated. First, I defined a model, activated USE_I18N = True in settings.py, run django-admin makemessages -l ru. No entries are created by default for model names and attributes. Grepping in the Django source code I found: $ ack "Select %s to change" contrib/admin/views/main.py 70: self.title = (self.is_popup and ugettext('Select %s') % force_unicode(self.opts.verbose_name) or ugettext('Select %s to change') % force_unicode(self.opts.verbose_name)) So the verbose_name meta property seems to play some role here. Tried to use it: class Order(models.Model): subject = models.CharField(max_length=150) description = models.TextField() class Meta: verbose_name = _('order') Now the updated po file contains msgid 'order' that can be translated. So I put the translation in. Unfortunately running the admin pages show the same mix of "???????? order ??? ?????????". I'm currently using Django 1.1.1. Could somebody point me to the relevant documentation? Because google can not. ;-) In the mean time I'll dig deeper into the django source code...

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  • how do you set the admin password on openldap 2.4

    - by dingfelder
    I am getting started with openLdap 2.4 and am having a bit of trouble, all the examples I see seem to refer to previous versions which used the text config file slapd.conf but from what I see on discussions about v2.4, this has been deprecated. I thought prehaps I needed to add a user, and log in as them but when I try and run an ldapadd command, I get a prompt to enter a password: Enter LDAP Password: ldap_bind: Invalid credentials (49) Notes: I installed openldap server via yum (in fedora 15), and have installed phpldapadminbut also can try things on the command line if anyone has suggestions. After installing and starting I get the following response from a search: # ldapsearch -x -b '' -s base '(objectclass=*)' namingContexts # extended LDIF # LDAPv3 # base <> with scope baseObject # filter: (objectclass=*) # requesting: namingContexts dn: namingContexts: dc=my-domain,dc=com # search result search: 2 result: 0 Success # numResponses: 2 # numEntries: 1 I am glad to remove and reinstall the server if that helps, can anyone provide a link to tips that works for version 2.4 for a new setup?

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  • Fed up of Server Admin - Want career change guidance

    - by JB04
    Hi All, I am SA in top level MNC and what I liked turned out to be my most disliked. I feel that I am capable of doing more than what I am doing at present. This 1 hour , 2 hour SLA is not my kind. I wanna get a better life.. The rotational shift is also something I am hating these days. Awkward shifts and too many process to follow. I have 3 years of Experience. I dont wanna waste this 3 yrs of experience I wanna get into OS developer or kind of so that this three years of experience is not wasted !! Please help me out

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  • Django admin site populated combo box based on imput

    - by user292652
    hi i have to following model class Match(models.Model): Team_one = models.ForeignKey('Team', related_name='Team_one') Team_two = models.ForeignKey('Team', related_name='Team_two') Stadium = models.CharField(max_length=255, blank=True) Start_time = models.DateTimeField(auto_now_add=False, auto_now=False, blank=True, null=True) Rafree = models.CharField(max_length=255, blank=True) Judge = models.CharField(max_length=255, blank=True) Winner = models.ForeignKey('Team', related_name='winner', blank=True) updated = models.DateTimeField('update date', auto_now=True ) created = models.DateTimeField('creation date', auto_now_add=True ) def save(self, force_insert=False, force_update=False): pass @models.permalink def get_absolute_url(self): return ('view_or_url_name') class MatchAdmin(admin.ModelAdmin): list_display = ('Team_one','Team_two', 'Winner') search_fields = ['Team_one','Team_tow'] admin.site.register(Match, MatchAdmin) i was wondering is their a way to populated the winner combo box once the team one and team two is selected in admin site ?

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  • Django: UserProfile with Unique Foreign Key in Django Admin

    - by lazerscience
    Hi, I have extended Django's User Model using a custom user profile called UserExtension. It is related to User through a unique ForeignKey Relationship, which enables me to edit it in the admin in an inline form! I'm using a signal to create a new profile for every new user: def create_user_profile(sender, instance, created, **kwargs): if created: try: profile, created = UserExtension.objects.get_or_create(user=instance) except: pass post_save.connect(create_user_profile, sender=User) (as described here for example: http://stackoverflow.com/questions/44109/extending-the-user-model-with-custom-fields-in-django) The problem is, that, if I create a new user through the admin, I get an IntegritiyError on saving "column user_id is not unique". It doesnt seem that the signal is called twice, but i guess the admin is trying to save the profile AFTERWARDS? But I need the creation through signal if I create a new user in other parts of the system!

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  • django-admin formfield_for_* change default value per/depending on instance

    - by Nick Ma.
    Hi, I'm trying to change the default value of a foreignkey-formfield to set a Value of an other model depending on the logged in user. But I'm racking my brain on it... This: Changing ForeignKey’s defaults in admin site would an option to change the empty_label, but I need the default_value. #Now I tried the following without errors but it didn't had the desired effect: class EmployeeAdmin(admin.ModelAdmin): ... def formfield_for_foreignkey(self, db_field, request=None, **kwargs): formfields= super(EmployeeAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) if request.user.is_superuser: return formfields if db_field.name == "company": #This is the RELEVANT LINE kwargs["initial"] = request.user.default_company return db_field.formfield(**kwargs) admin.site.register(Employee, EmployeeAdmin) ################################################################## # REMAINING Setups if someone would like to know it but i think # irrelevant concerning the problem ################################################################## from django.contrib.auth.models import User, UserManager class CompanyUser(User): ... objects = UserManager() company = models.ManyToManyField(Company) default_company= models.ForeignKey(Company, related_name='default_company') #I registered the CompanyUser instead of the standard User, # thats all up and working ... class Employee(models.Model): company = models.ForeignKey(Company) ... Hint: kwargs["default"] ... doesn't exist. Thanks in advance, Nick

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  • Django Admin interface with pickled set

    - by Rosarch
    I have a model that has a pickled set of strings. (It has to be pickled, because Django has no built in set field, right?) class Foo(models.Model): __bar = models.TextField(default=lambda: cPickle.dumps(set()), primary_key=True) def get_bar(self): return cPickle.loads(str(self.__bar)) def set_bar(self, values): self.__bar = cPickle.dumps(values) bar = property(get_bar, set_bar) I would like the set to be editable in the admin interface. Obviously the user won't be working with the pickled string directly. Also, the interface would need a widget for adding/removing strings from a set. What is the best way to go about doing this? I'm not super familiar with Django's admin system. Do I need to build a custom admin widget or something? Update: If I do need a custom widget, this looks helpful: http://www.fictitiousnonsense.com/archives/22

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  • Django admin site auto populate combo box based on input

    - by user292652
    hi i have to following model class Match(models.Model): Team_one = models.ForeignKey('Team', related_name='Team_one') Team_two = models.ForeignKey('Team', related_name='Team_two') Stadium = models.CharField(max_length=255, blank=True) Start_time = models.DateTimeField(auto_now_add=False, auto_now=False, blank=True, null=True) Rafree = models.CharField(max_length=255, blank=True) Judge = models.CharField(max_length=255, blank=True) Winner = models.ForeignKey('Team', related_name='winner', blank=True) updated = models.DateTimeField('update date', auto_now=True ) created = models.DateTimeField('creation date', auto_now_add=True ) def save(self, force_insert=False, force_update=False): pass @models.permalink def get_absolute_url(self): return ('view_or_url_name') class MatchAdmin(admin.ModelAdmin): list_display = ('Team_one','Team_two', 'Winner') search_fields = ['Team_one','Team_tow'] admin.site.register(Match, MatchAdmin) i was wondering is their a way to populated the winner combo box once the team one and team two is selected in admin site ?

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  • Using Django Admin for a custom database solution

    - by Prashanth Ellina
    A client wants to have a simple intranet application to manage his process. He runs a Quarry and wishes to track number of loads delivered per day and associated activities. Since I knew about Django's excellent Admin interface, I figured I could define the "Schema" in models.py and have Django Admin generate the forms. I did exactly that and the result is not bad at all. I've been able to customize the look and feel to suit the client's taste. Some questions: Is Django Admin the right choice for such a use-case? Will I run to problems in the future due to flexibility of the framework? Is there a better framework out there specifically designed for this use-case (general Database management for small businesses)? I prefer ones written in Python since I can hack it up to customize. Thanks!

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