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  • How to customize a many-to-many inline model in django admin

    - by Jonathan
    I'm using the admin interface to view invoices and products. To make things easy, I've set the products as inline to invoices, so I will see the related products in the invoice's form. As you can see I'm using a many-to-many relationship. In models.py: class Product(models.Model): name = models.TextField() price = models.DecimalField(max_digits=10,decimal_places=2) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) In admin.py: class ProductInline(admin.StackedInline): model = Invoice.products.through class InvoiceAdmin(admin.ModelAdmin): inlines = [FilteredApartmentInline,] admin.site.register(Product, ProductAdmin) The problem is that django presents the products as a table of drop down menus (one per associated product). Each drop down contains all the products listed. So if I have 5000 products and 300 are associated with a certain invoice, django actually loads 300x5000 product names. Also the table is not aesthetic. How can I change it so that it'll just display the product's name in the inline table? Which form should I override, and how?

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  • ValueError with multi-table inheritance in Django Admin

    - by jorde
    I created two new classes which inherit model Entry: class Entry(models.Model): LANGUAGE_CHOICES = settings.LANGUAGES language = models.CharField(max_length=2, verbose_name=_('Comment language'), choices=LANGUAGE_CHOICES) user = models.ForeignKey(User) country = models.ForeignKey(Country, null=True, blank=True) created = models.DateTimeField(auto_now=True) class Comment(Entry): comment = models.CharField(max_length=2000, blank=True, verbose_name=_('Comment in English')) class Discount(Entry): discount = models.CharField(max_length=2000, blank=True, verbose_name=_('Comment in English')) coupon = models.CharField(max_length=2000, blank=True, verbose_name=_('Coupon code if needed')) After adding these new models to admin via admin.site.register I'm getting ValueError when trying to create a comment or a discount via admin. Adding entries works fine. Error msg: ValueError at /admin/reviews/discount/add/ Cannot assign "''": "Discount.discount" must be a "Discount" instance. Request Method: GET Request URL: http://127.0.0.1:8000/admin/reviews/discount/add/ Exception Type: ValueError Exception Value: Cannot assign "''": "Discount.discount" must be a "Discount" instance. Exception Location: /Library/Python/2.6/site-packages/django/db/models/fields/related.py in set, line 211 Python Executable: /usr/bin/python Python Version: 2.6.1

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  • Django admin.py missing field error

    - by user782400
    When I include 'caption', I get an error saying EntryAdmin.fieldsets[1][1]['fields']' refers to field 'caption' that is missing from the form In the admin.py; I have imported the classes from joe.models import Entry,Image Is that because my class from models.py is not getting imported properly ? Need help in resolving this issue. Thanks. models.py class Image(models.Model): image = models.ImageField(upload_to='joe') caption = models.CharField(max_length=200) imageSrc = models.URLField(max_length=200) user = models.CharField(max_length=20) class Entry(models.Model): image = models.ForeignKey(Image) mimeType = models.CharField(max_length=20) name = models.CharField(max_length=200) password = models.URLField(max_length=50) admin.py class EntryAdmin(admin.ModelAdmin): fieldsets = [ ('File info', {'fields': ['name','password']}), ('Upload image', {'fields': ['image','caption']})] list_display = ('name', 'mimeType', 'password') admin.site.register(Entry, EntryAdmin) admin.site.register(Image)

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  • Dajano admin site foreign key fields

    - by user292652
    hi i have the following models setup class Player(models.Model): #slug = models.slugField(max_length=200) Player_Name = models.CharField(max_length=100) Nick = models.CharField(max_length=100, blank=True) Jersy_Number = models.IntegerField() Team_id = models.ForeignKey('Team') Postion_Choices = ( ('M', 'Manager'), ('P', 'Player'), ) Poistion = models.CharField(max_length=1, blank=True, choices =Postion_Choices) Red_card = models.IntegerField( blank=True, null=True) Yellow_card = models.IntegerField(blank=True, null=True) Points = models.IntegerField(blank=True, null=True) #Pic = models.ImageField(upload_to=path/for/upload, height_field=height, width_field=width, max_length=100) class PlayerAdmin(admin.ModelAdmin): list_display = ('Player_Name',) search_fields = ['Player_Name',] admin.site.register(Player, PlayerAdmin) class Team(models.Model): """Model docstring""" #slug = models.slugField(max_length=200) Team_Name = models.CharField(max_length=100,) College = models.CharField(max_length=100,) Win = models.IntegerField(blank=True, null=True) Loss = models.IntegerField(blank=True, null=True) Draw = models.IntegerField(blank=True, null=True) #logo = models.ImageField(upload_to=path/for/upload, height_field=height, width_field=width, max_length=100) class Meta: pass #def __unicode__(self): # return Team_Name #def save(self, force_insert=False, force_update=False): # pass @models.permalink def get_absolute_url(self): return ('view_or_url_name') class TeamAdmin(admin.ModelAdmin): list_display = ('Team_Name',) search_fields = ['Team_Name',] admin.site.register(Team, TeamAdmin) my question is how do i get to the admin site to show Team_name in the add player form Team_ID field currently it is only showing up as Team object in the combo box

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  • How to deal with multiple sub-type of one super-type in Django admin

    - by Henri
    What would be the best solution for adding/editing multiple sub-types. E.g a super-type class Contact with sub-type class Client and sub-type class Supplier. The way shown here works, but when you edit a Contact you get both inlines i.e. sub-type Client AND sub-type Supplier. So even if you only want to add a Client you also get the fields for Supplier of vice versa. If you add a third sub-type , you get three sub-type field groups, while you actually only want one sub-type group, in the mentioned example: Client. E.g.: class Contact(models.Model): contact_name = models.CharField(max_length=128) class Client(models.Model): contact = models.OneToOneField(Contact, primary_key=True) user_name = models.CharField(max_length=128) class Supplier(models.Model): contact.OneToOneField(Contact, primary_key=True) company_name = models.CharField(max_length=128) and in admin.py class ClientInline(admin.StackedInline): model = Client class SupplierInline(admin.StackedInline): model = Supplier class ContactAdmin(admin.ModelAdmin): inlines = (ClientInline, SupplierInline,) class ClientAdmin(admin.ModelAdmin): ... class SupplierAdmin(admin.ModelAdmin): ... Now when I want to add a Client, i.e. only a Client I edit Contact and I get the inlines for both Client and Supplier. And of course the same for Supplier. Is there a way to avoid this? When I want to add/edit a Client that I only see the Inline for Client and when I want to add/edit a Supplier that I only see the Inline for Supplier, when adding/editing a Contact? Or perhaps there is a different approach. Any help or suggestion will be greatly appreciated.

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  • Cannot Delete Shortcut from Desktop because I need Admin Permissions even though I am Admin

    - by DavidB
    Installing the new Malwarebytes 2.0 put a shortcut on my desktop that I want to remove, but dragging it to the recycle bin shows this message: I am an administrator on this computer, so normally clicking "Continue" solves the problem, but it didn't work here. Instead, I got this message. How can I resolve this? I have tried using the built in super administrator account to remove it, but that does not work either.

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  • xampp admin page access forbidden

    - by Vihaan Verma
    I m new to apache world ! I read some docs online to setup virtual host . Which works fine ! Here are the content of httpd-vhosts.conf file <Directory C:/vhosts> Order Deny,Allow Allow from all </Directory> NameVirtualHost *:80 <VirtualHost *:80> DocumentRoot "C:/htdocs" ServerName localhost </VirtualHost> <VirtualHost *:80> DocumentRoot "C:/vhosts/phpdw" ServerName phpdw </VirtualHost> But now when I access the xampp control panel and try accessing the apache admin page I get access defined eror (403) . My guess is that there needs to be some more configuration in this file to allow access to localhost. I could not find anything relevant . Thanks

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  • nginx not serving admin static files?

    - by toto_tico
    First, I want to clarify that this error is just for the admin static files. This means my problem is specific just to the static files that corresponds to the Django admin. The rest of the static files are working perfect. Basically my problem is that for some reason I cannot access those admin static files with the ngix server. It works perfect with the micro server of Django and the collect static is doing its job. This means it is putting the files on the expected place in the static folder. The urls are correct but I cannot even access the admin static files directly, but the others I can. So, for example, I am able to access this url (copying it in the browser): myserver.com:8080/static/css/base/base.css but i am not able to access this other url (copying it in the browser): myserver.com:8080/static/admin/css/admin.css I also tried to copy the admin/ directory structure into other_admin_directory_name/. Then I can access myserver.com:8080/static/other_admin_directory_name/css/admin.css Then, it works. So, I checked permissions and everything is fine. I tried to change ADMIN_MEDIA_PREFIX = '/static/admin/' to ADMIN_MEDIA_PREFIX = '/static/other_admin_directory_name/', it doesn't work. This a mistery in itself that I am exploring but still no luck. Finally, and it seems to be an important clue: I tried to copy the admin/ directory structure into admin_and_then_any_suffix/. Then I cannot access myserver.com:8080/static/admin_and_then_any_suffix//css/admin.css So, if the name of the directory starts with admin (for example administration or admin2) it doesn't work. * added thanks to sarnold observation ** the problem seems to be in the nginx configuration file /etc/nginx/sites-available/mysite location /static/admin { alias /home/vl3/.virtualenvs/vl3/lib/python2.7/site-packages/django/contrib/admin/media/; }

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  • Prepopulating inlines based on the parent model in the Django Admin

    - by Alasdair
    I have two models, Event and Series, where each Event belongs to a Series. Most of the time, an Event's start_time is the same as its Series' default_time. Here's a stripped down version of the models. #models.py class Series(models.Model): name = models.CharField(max_length=50) default_time = models.TimeField() class Event(models.Model): name = models.CharField(max_length=50) date = models.DateField() start_time = models.TimeField() series = models.ForeignKey(Series) I use inlines in the admin application, so that I can edit all the Events for a Series at once. If a series has already been created, I want to prepopulate the start_time for each inline Event with the Series' default_time. So far, I have created a model admin form for Event, and used the initial option to prepopulate the time field with a fixed time. #admin.py ... import datetime class OEventInlineAdminForm(forms.ModelForm): start_time = forms.TimeField(initial=datetime.time(18,30,00)) class Meta: model = OEvent class EventInline(admin.TabularInline): form = EventInlineAdminForm model = Event class SeriesAdmin(admin.ModelAdmin): inlines = [EventInline,] I am not sure how to proceed from here. Is it possible to extend the code, so that the initial value for the start_time field is the Series' default_time?

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  • Rate limiting Django admin login with Nginx to prevent dictionary attack

    - by shreddies
    I'm looking into the various methods of rate limiting the Django admin login to prevent dictionary attacks. One solution is explained here: simonwillison.net/2009/Jan/7/ratelimitcache/ However, I would prefer to do the rate limiting at the web server side, using Nginx. Nginx's limit_req module does just that - allowing you to specify the maximum number of requests per minute, and sending a 503 if the user goes over: http://wiki.nginx.org/NginxHttpLimitReqModule Perfect! I thought I'd cracked it until I realised that Django admin's login page is not in a consistent place, eg /admin/blah/ gives you a login page at that URL, rather than bouncing to a standard login page. So I can't match on the URL. Can anyone think of another way to know that the admin page was being displayed (regexp the response HTML?)

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  • Django Admin drop down combobox and assigned values

    - by Daniel Garcia
    I have several question for the Django Admin feature. Im kind of new in Django so im not sure how to do it. Basically what Im looking to do is when Im adding information on the model. Some of the fields i want them to be drop-downs and maybe combo-boxes with AutoCompleteMode. Also looking for some fields to have the same information, for example if i have a datatime field I want that information to feed the fields day, month and year from hoti.hotiapp.models import Occurrence from django.contrib import admin class MyModelAdmin(admin.ModelAdmin): exclude = ['reference',] admin.site.register(Occurrence, MyModelAdmin) Anything helps Thanks in advance

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  • Tying in to Django Admin's Model History

    - by akdom
    The Setup: I'm working on a Django application which allows users to create an object in the database and then go back and edit it as much as they desire. Django's admin site keeps a history of the changes made to objects through the admin site. The Question: How do I hook my application in to the admin site's change history so that I can see the history of changes users make to their "content" ?

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  • Django Admin: not seeing any app (permission problem?)

    - by Facundo
    I have a site with Django running some custom apps. I was not using the Django ORM, just the view and templates but now I need to store some info so I created some models in one app and enabled the Admin. The problem is when I log in the Admin it just says "You don't have permission to edit anything", not even the Auth app shows in the page. I'm using the same user created with syncdb as a superuser. In the same server I have another site that is using the Admin just fine. Using Django 1.1.0 with Apache/2.2.10 mod_python/3.3.1 Python/2.5.2, with psql (PostgreSQL) 8.1.11 all in Gentoo Linux 2.6.23 Any ideas where I can find a solution? Thanks a lot. UPDATE: It works from the development server. I bet this has something to do with some filesystem permission but I just can't find it. UPDATE2: vhost configuration file: <Location /> SetHandler python-program PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE gpx.settings PythonDebug On PythonPath "['/var/django'] + sys.path" </Location> UPDATE 3: more info /var/django/gpx/init.py exists and is empty I run python manage.py from /var/django/gpx directory The site is GPX, one of the apps is contable and lives in /var/django/gpx/contable the user apache is webdev group and all these directories and files belong to that group and have rw permission UPDATE 4: confirmed that the settings file is the same for apache and runserver (renamed it and both broke) UPDATE 5: /var/django/gpx/contable/init.py exists This is the relevan part of urls.py: urlpatterns = patterns('', (r'^admin/', include(admin.site.urls)), ) urlpatterns += patterns('gpx', (r'^$', 'menues.views.index'), (r'^adm/$', 'menues.views.admIndex'),

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  • django: displaying group users count in admin

    - by gruszczy
    I would like to change admin for a group, so it would display how many users are there in a certain group. I'd like to display this in the view showing all groups, the one before you enter admin for certain group. Is it possible? I am talking both about how to change admin for a group and how to add function to list_display.

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  • How do I reinstate my admin user privileges to global read/write

    - by Matt
    I am running Ubuntu 12.04 LTS. I only have the one user which I created when I installed Ubuntu. Everything has been fine - love it - until I updated a software package recently from the command line using sudo (not gksudo). I was having a little bother which did not make sense to me and in a fluff changed my user read/write privileges through the GUI (not even clear how I got there!). After restart I was stuck in a login loop - using the right login password but kept getting looped back to the login and could only login as Guest. I could still login with my user/password via ctrl + alt + f1 Eventually I was able to login again at start up. Not sure exactly what it was I changed that worked but it was one of/or a combination of installing latest security updates, changing login manager from LightDM to DGM and back again, removing the ICE/Xauthority and chown user. Current dilemma is my primary admin user privileges were read only. In the command line ls -ls /home/user returned this value: drwx------ 48 username username 20480 I have since changed this using sudo chmod 0755 /home/username (from my limited understanding 755 should return my user privileges to their original read/write glory). ls -ld /home/user currently shows my user privileges as: drwxr-xr-x 48 username username 20480 I still seem to have only read access permissions. I've been through lots of threads (and the help file) that talk about creating new users/groups permissions etc. but specific info on returning my existing global/admin/primary users privileges to what they were when I first created that user - baffling me. I feel this is something really simple I'm just not getting it. Please help! sudo mount /dev/sda1 on / type ext4 (rw,errors=remount-ro) proc on /proc type proc (rw,noexec,nosuid,nodev) sysfs on /proc type sysfs (rw,noexec,nosuid,nodev) none on /sys/fs/fuse/connections type fusect1 (rw) none on /sys/kernel/debug type debugfs (rw) none on /sys/kernel/security type securityfs (rw) udev on /dev type devtmpfs (rw,mode=07pe tmpfs55) devpts on /dev/pts type devpts (rw,noexec,nosuid,gid=5,mode=0620) tmpfs on /run type tmpfs (rw,noexec,nosuid,size=10%,mode=0755) none on /run/lock type tmpfs (rw, ,nosuid,nodev,size=5242880 none on /run/shm type tmpfs (rw,nosuid,nodev) gvfs-fuse-daemon on /home/meng/.gvfs type fuse.gvfs-fuse-daemon (rw,nosuid,nodev,user=meng) none on /tmp/guest-1R2Fi5 type tmpsf (rw,mode=700)

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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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