Search Results

Search found 6239 results on 250 pages for 'rational admin'.

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 4/250 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • 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?

    Read the article

  • 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

    Read the article

  • 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 ?

    Read the article

  • 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!

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • 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 ?

    Read the article

  • 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!

    Read the article

  • django admin app error (Model with property field): global name 'full_name' is not defined

    - by rxin
    This is my model: class Author(models.Model): first_name = models.CharField(max_length=200) last_name = models.CharField(max_length=200) middle_name = models.CharField(max_length=200, blank=True) def __unicode__(self): return full_name def _get_full_name(self): "Returns the person's full name." if self.middle_name == '': return "%s %s" % (self.first_name, self.last_name) else: return "%s %s %s" % (self.first_name, self.middle_name, self.last_name) full_name = property(_get_full_name) Everything is fine except when I go into admin interface, I see TemplateSyntaxError at /bibbase2/admin/bibbase2/author/ Caught an exception while rendering: global name 'full_name' is not defined It seems like the built-in admin app doesn't work with a property field. Is there something wrong with my code?

    Read the article

  • group inlines in django admin

    - by pablo
    Hi I have a two models, Model1 and Model2. Model2 has a FK to Model1 and FK to iteself. In the admin I show Model2 as inlines in Model1 change_form. I want to modify the way the inlines are shown in the admin. I need to group all the instances that have the same parent_model2 and display them as a readonly field with a string of 'childs' in the parent Model2 instance. I know how to use itertools.groupby (or the django version) but don't know how to do it in the admin. What should I override to be able to iterate over all the Model2 instances, group them by parent, add children to the parent and remove children from the inlines? class Model1(models.Model): name = models.CharField() class Model2(models.Model): name = models.CharField() fk_model1 = models.ForeignKey('self', blank=True, null=True) parent_model2 = models.ForeignKey('self', blank=True, null=True) Thanks

    Read the article

  • Filtering foreign keys with AJAX in Django admin

    - by cnobile
    I have most of this figured out already. I have AJAX returning the region/state/province when a country is selected. The correct foreign key is saved to the database, however, when the record is viewed afterwards the selected state is not shown in the select nor are any states for the selected country. I understand why this is happening as, the admin view is not aware of the relation between the state and the country. So here is the question. Is there a hook in the admin view that will allow me to load the correct states for the country and set the selected attribute on the option in the select tag? Or how can I override the admin view for any forms that require the country and region/state/province set? I am using jQuery and Djando-1.1. Thanks

    Read the article

  • Inlines in Django Admin

    - by Oli
    I have two models, Order and UserProfile. Each Order has a ForeignKey to UserProfile, to associate it with that user. On the django admin page for each Order, I'd like to display the UserProfile associated with it, for easy processing of information. I have tried inlines: class UserInline(admin.TabularInline): model = UserProfile class ValuationRequestAdmin(admin.ModelAdmin): list_display = ('address1', 'address2', 'town', 'date_added') list_filter = ('town', 'date_added') ordering = ('-date_updated',) inlines = [ UserInline, ] But it complains that UserProfile "has no ForeignKey to" Order - which it doesn't, it's the other way around. Is there a way to do what I want?

    Read the article

  • Django Admin: Many-to-Many listbox doesn't show up with a through parameter

    - by NP
    Hi All, I have the following models: class Message(models.Model): date = models.DateTimeField() user = models.ForeignKey(User) thread = models.ForeignKey('self', blank=True, null=True) ... class Forum(models.Model): name = models.CharField(max_length=24) messages = models.ManyToManyField(Message, through="Message_forum", blank=True, null=True) ... class Message_forum(models.Model): message = models.ForeignKey(Message) forum = models.ForeignKey(Forum) status = models.IntegerField() position = models.IntegerField(blank=True, null=True) tags = models.ManyToManyField(Tag, blank=True, null=True) In the admin site, when I go to add/change a forum, I don't see the messages listbox as you'd expect. However, it shows up if I remove the 'through' parameter in the ManyToManyField declaration. What's up with that? I've registered all three models (plus Tag) to the admin site in admin.py. TIA

    Read the article

  • Can not login Magento admin page after move to my localhost

    - by Xinrui Ma
    I just move my Magento store to my localhost environment for testing use, I also using Git to maintain code, but after I move all the files to my local environment, I can't login my admin page, but I can still see my frontend pages, and the git, the database, seems works well. When I type a wrong admin/password to my admin page, it still gives me "Invalid password". But when I enter the right one, it just refresh the page and stay at the login page, nothing happens. Does anyone has met this problem before? Has any ideas? Thanks in advance!

    Read the article

  • Sorting related objects in the Django Admin form interface

    - by Carver
    I am looking to sort the related objects that show up when editing an object using the admin form. So for example, I would like to take the following object: class Person(models.Model): first_name = models.CharField( ... ) last_name = models.CharField( ... ) hero = models.ForeignKey( 'self', null=True, blank=True ) and edit the first name, last name and hero using the admin interface. I want to sort the objects as they show up in the drop down by last name, first name (ascending). How do I do that? Context I'm using Django v1.1. I started by looking for help in the django admin docs, but didn't find the solution As you can see in the example, the foreign key is pointing to itself, but I expect it would be the same as pointing to a different model object. Bonus points for being able to filter the related objects, too (eg~ only allow selecting a hero with the same first name)

    Read the article

  • Django admin, hide a model

    - by Hellnar
    Hello At the root page of the admin site where registered models appear, I want to hide several models that are registered to the Django admin. If I directly unregister those, I am not able to add new records as the add new symbol "+" dissapears. How can this be done ?

    Read the article

  • Edit/show Primary Key in Django Admin

    - by emcee
    It appears Django hides fields that are flagged Primary Key from being displayed/edited in the Django admin interface. Let's say I'd like to input data in which I may or may not want to specify a primary key. How would I go about displaying primary keys in Django-admin, and how could I make specifying it optional? Many thanks in advance, beloved hive-mind.

    Read the article

  • django admin site make CharField a PasswordInput

    - by Paul
    I have a Django site in which the site admin inputs their Twitter Username/Password in order to use the Twitter API. The Model is set up like this: class TwitterUser(models.Model): screen_name = models.CharField(max_length=100) password = models.CharField(max_length=255) def __unicode__(self): return self.screen_name I need the Admin site to display the password field as a password input, but can't seem to figure out how to do it. I have tried using a ModelAdmin class, a ModelAdmin with a ModelForm, but can't seem to figure out how to make django display that form as a password input...

    Read the article

  • Using ckEditor on selective text areas in django admin forms

    - by Rahul
    Hi, I want to apply ckeditor on specific textarea in django admin form not on all the text areas. Like snippet below will apply ckeditor on every textarea present on django form: class ProjectAdmin(admin.ModelAdmin): formfield_overrides = {models.TextField: {'widget': forms.Textarea(attrs={'class':'ckeditor'})}, } class Media: js = ('ckeditor/ckeditor.js',) but i want it on a specific textarea not on every textarea.

    Read the article

  • display one-to-many relationship for a model in Django admin (list mode)

    - by theactiveactor
    In Django admin site, when listing all the objects for a given model, I know we can customize which columns get displayed for a ModelA via list_display Say that ModelA has a one-to-many relationship with ModelB. I would like to add another column on the listing page for ModelA, where each entry is a URL pointing to all objects of ModelB having a foreign key relationship on corresponding instance of Model A in that row. How can I achieve this customization with the admin app?

    Read the article

  • Django admins disappearing from the admin index

    - by btol45
    We're running a Django website with rough 45 install Django admin classes. The handler is mod_fastcgi. Every once in a while about half the admins disappear from /admin/ screen. Touching the production.fcgi file restores everything to normal, but we have yet to determine the underling cause. Any thoughts on what the underlying issue might be?

    Read the article

  • Foreign keys in django admin list display

    - by Olivier
    If a django model contains a foreign key field, and if that field is shown in list mode, then it shows up as text, instead of displaying a link to the foreign object. Is it possible to automatically display all foreign keys as links instead of flat text? (of course it is possible to do that on a field by field basis, but is there a general method?) Example: class Author(models.Model): ... class Post(models.Model): author = models.ForeignKey(Author) Now I choose a ModelAdmin such that the author shows up in list mode: class PostAdmin(admin.ModelAdmin): list_display = [..., 'author',...] Now in list mode, the author field will just use the __unicode__ method of the Author class to display the author. On the top of that I would like a link pointing to the url of the corresponding author in the admin site. Is that possible? Manual method: For the sake of completeness, I add the manual method. It would be to add a method author_link in the PostAdmin class: def author_link(self, item): return '<a href="../some/path/%d">%s</a>' % (item.id, unicode(item)) author_link.allow_tags = True That will work for that particular field but that is not what I want. I want a general method to achieve the same effect. (One of the problems is how to figure out automatically the path to an object in the django admin site.)

    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 | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >