Search Results

Search found 58396 results on 2336 pages for 'system admin and developer'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 7/2336 | < Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >

  • 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)

    Read the article

  • 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

    Read the article

  • What should a Java/SOA developer be able to do?

    - by Regular Joe
    Hello community. I got assigned the task to list the activities a Java Developer should be able to perform and create an estimate about the time it would take. I've came up with the following: S = Small complexity M = Medium complexity H = High complexity 1d = 1 day Create JDBC CRUD backend ( S=1d, M=5d, H=10d ) Create JSP/Servlet frontend for a CRUD app ( S=1d, M=10d, H=20d ) Create Swing desktop frontend ( S=1d, M=15d, H=30d) Create ORM based CRUD ... Create Webapp fronend with webframework ... This is thought for a Java "enterprise" developer. The other profile I have is SOA Developer, but I could not pass beyond: Create webservice ( S=.5d, M=2d, H=7d ) Q.- What other activities should a Java Developer be able to do? Q.- What activities should a SOA Developer be able to do? Please, help me with this, I know this is in the limit of the kind of questions that could be asked here, but I really need a little push on this, and I don't want to go to Yahoo Answers for this.

    Read the article

  • 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.

    Read the article

  • From Developer to Web Developer to Web Designer

    - by leftbrainlogic
    Is it possible for a fairly experienced Java Developer to transition to being a Web Developer and then to Web Designer. I guess what I'm asking is - assume you have (Java) developer of above average aptitute - is it possible for that developer to acquire web development skills that will enable him/her to develop small business websites without the need to hire outside skills. If so, where does one start on the path to becoming a Web Developer/Designer? Skills required? Tools used?

    Read the article

  • DC: Mac Developer vs iPhone Developer

    - by Khou
    http://developer.apple.com/programs/ so whats the difference between Mac Developer and iPhone Developer membership? If you signup to Mac Developer you can download the iPhone SDK anyways? so why would you sign to iphone developer?

    Read the article

  • Object-Oriented Operating System

    - by nmagerko
    As I thought about writing an operating system, I came across a point that I really couldn't figure out on my own: Can an operating system truly be written in an Object-Oriented Programming (OOP) Language? Being that these types of languages do not allow for direct accessing of memory, wouldn't this make it impossible for a developer to write an entire operating system using only an OOP Language? Take, for example, the Android Operating System that runs many phones and some tablets in use around the world. I believe that this operating system uses only Java, an Object-Oriented language. In Java, I have been unsuccessful in trying to point at and manipulate a specific memory address that the run-time environment (JRE) has not assigned to my program implicitly. In C, C++, and other non-OOP languages, I can do this in a few lines. So this makes me question whether or not an operating system can be written in an OOP, especially Java. Any counterexamples or other information is appreciated.

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • System.Threading.ThreadAbortException executing WCF service

    - by SURESH GIRIRAJAN
    In one of our prod server we recently ran into issue when we went and update the web.config and try to browse the service. We started seeing the service was not responding and getting the following warning in the application log. Our service is WCF service, BizTalk orchestration exposed as service. We have other prod server where we never ran into this issue, so what’s different with this server. After going thru lot of forum and came up on some Microsoft service pack and hot fix which related to FCN. But I don’t want to apply any patch on this server then we need to do on all the other servers too. So solution is simple, I dropped the existing website, created a new site with different name with updated web.config browse the service. Then dropped that site and recreate the original web site and it worked fine without any issue. Event Viewer:  Event Type:        Warning Event Source:    ASP.NET 2.0.50727.0 Event Category:                Web Event Event ID:              1309 Date:                     6/6/2011 Time:                    5:41:42 PM User:                     N/A Computer:          PRODP02 Description: Event code: 3005 Event message: An unhandled exception has occurred. Event time: 6/6/2011 5:41:42 PM Event time (UTC): 6/6/2011 9:41:42 PM Event ID: a71769f42b304355a58c482bfec267f2 Event sequence: 3 Event occurrence: 1 Event detail code: 0  Application information:     Application domain: /LM/W3SVC/518296899/ROOT/PortArrivals-2-129518698821558995     Trust level: Full     Application Virtual Path: /TESTSVC     Application Path: D:\inetpub\wwwroot\RFID\TESTSVC\     Machine name: PRODP02  Process information:     Process ID: 8752     Process name: w3wp.exe     Account name: domain\BizTalk_Svc_Hostlso  Exception information:     Exception type: ThreadAbortException     Exception message: Thread was being aborted.  Request information:     Request URL: http://localhost:81/TESTSVC/TESTSVCS.svc     Request path: /TESTSVC/TESTSVCS.svc     User host address: 127.0.0.1     User:      Is authenticated: False     Authentication Type:      Thread account name: domain\BizTalk_Svc_Hostlso  Thread information:     Thread ID: 22     Thread account name: domain\BizTalk_Svc_Hostlso     Is impersonating: False     Stack trace:    at System.Web.HttpApplication.ExecuteStep(IExecutionStep step, Boolean& completedSynchronously)    at System.Web.HttpApplication.ApplicationStepManager.ResumeSteps(Exception error)  at System.Web.HttpApplication.System.Web.IHttpAsyncHandler.BeginProcessRequest(HttpContext context, AsyncCallback cb, Object extraData)    at System.Web.HttpRuntime.ProcessRequestInternal(HttpWorkerRequest wr)  <Description>Handling an exception.</Description> <AppDomain>/LM/W3SVC/518296899/ROOT/TESTSVC-6-129518741899334691</AppDomain> <Exception> <ExceptionType>System.Threading.ThreadAbortException, mscorlib, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089</ExceptionType> <Message>Thread was being aborted.</Message> <StackTrace> at System.Threading.Monitor.Enter(Object obj) at System.ServiceModel.ServiceHostingEnvironment.HostingManager.EnsureServiceAvailable(String normalizedVirtualPath) at System.ServiceModel.ServiceHostingEnvironment.EnsureServiceAvailableFast(String relativeVirtualPath) at System.ServiceModel.Activation.HostedHttpRequestAsyncResult.HandleRequest() at System.ServiceModel.Activation.HostedHttpRequestAsyncResult.BeginRequest() at System.ServiceModel.Activation.HostedHttpRequestAsyncResult.OnBeginRequest(Object state) at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.WorkItem.Invoke2() at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.WorkItem.Invoke() at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.ProcessCallbacks() at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.CompletionCallback(Object state) at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.ScheduledOverlapped.IOCallback(UInt32 errorCode, UInt32 numBytes, NativeOverlapped* nativeOverlapped) at System.ServiceModel.Diagnostics.Utility.IOCompletionThunk.UnhandledExceptionFrame(UInt32 error, UInt32 bytesRead, NativeOverlapped* nativeOverlapped) </StackTrace> <ExceptionString>System.Threading.ThreadAbortException: Thread was being aborted.    at System.Threading.Monitor.Enter(Object obj)    at System.ServiceModel.ServiceHostingEnvironment.HostingManager.EnsureServiceAvailable(String normalizedVirtualPath)    at System.ServiceModel.ServiceHostingEnvironment.EnsureServiceAvailableFast(String relativeVirtualPath)    at System.ServiceModel.Activation.HostedHttpRequestAsyncResult.HandleRequest()    at System.ServiceModel.Activation.HostedHttpRequestAsyncResult.BeginRequest()    at System.ServiceModel.Activation.HostedHttpRequestAsyncResult.OnBeginRequest(Object state)    at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.WorkItem.Invoke2()    at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.WorkItem.Invoke()    at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.ProcessCallbacks()    at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.CompletionCallback(Object state)    at System.ServiceModel.Channels.IOThreadScheduler.CriticalHelper.ScheduledOverlapped.IOCallback(UInt32 errorCode, UInt32 numBytes, NativeOverlapped* nativeOverlapped)    at System.ServiceModel.Diagnostics.Utility.IOCompletionThunk.UnhandledExceptionFrame(UInt32 error, UInt32 bytesRead, NativeOverlapped* nativeOverlapped)</ExceptionString>

    Read the article

  • 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/; }

    Read the article

  • 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?

    Read the article

  • 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?)

    Read the article

  • 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

    Read the article

  • 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" ?

    Read the article

  • 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'),

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • What's the diffrence btw System property and system environment variable

    - by khue
    Hi all I am not clear about this. When I run a java App or run an Applet in applet viewer,( in the IDE environment), System.getProperty("java.class.path") give me the same as System.getenv("CLASSPATH"), which is the CLASSPATH env variable defined. But when I deploy my applet to webserver and access it from the same computer as a client, I get different result for the two (System.getProperty("java.class.path") only point to jre home and System.getenv("CLASSPATH") return null). And here is some other things that make me wonder: For the applet part, the env var JAVA_HOME, i get the same result when deploying the applet in a browser as well as Applet Viewer. And if I define myself a env variable at system level, and use getenv("envName") the result is null. Is there anyway I can define one and get it in my java program? Thanks a lot Regards K.

    Read the article

  • 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

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >