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  • How to change Windows admin password from guest user

    - by John Smiith
    How to gain access of admin account of Windows, I activated a guest user and I want to change the admin password from the command line. When I type: net user administrator password the response is System error 5 has occurred. Access is denied I am using winxp pro sp2 I am running this command from cdm.exe and I am running this command from guest user. I actually want to change my admin password from guest user.

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  • Problems Running Cherokee Web Server Admin - config_reader.c:249 - Parsing error

    - by Sebastian
    I'm running Cherokee web server 0.99.30 on (Ubuntu Hardy) and I have been having some issues getting the admin to run property. When I run sudo cherokee-admin -b Login: User: admin One-time Password: {password} Web Interface: URL: http://localhost:9090/ [20/11/2009 22:57:29.733] (error) config_reader.c:249 - Parsing error Cherokee Web Server 0.99.30 (Nov 20 2009): Listening on port ALL:9090, TLS disabled, IPv6 disabled, using epoll, 4096 fds system limit, max. 2041 connections, caching I/O, single thread When I go to the admin page I get a 503 Service Unavailable error page. Any idea about how I could fix this? Thanks

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  • how to login as admin in safemode?

    - by Peter
    My sister has forgotten her password to vista, however i have installed that system so should know the admin password. However I do not know how to log in as admin. i tried to press ctrl+alt+delete twice in safemode to switch to normal login mode, but its not working. I heard that admin account is by default turned off in vista, so it might not work.

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  • How to filter queryset in changelist_view in django admin?

    - by minder
    Let's say I have a site where Users can add Entries through admin panel. Each User has his own Category he is responsible for (each Category has an Editor assigned through ForeingKey/ManyToManyField). When User adds Entry, I limit the choices by using EntryAdmin like this: class EntryAdmin(admin.ModelAdmin): (...) def formfield_for_foreignkey(self, db_field, request, **kwargs): if db_field.name == 'category': if request.user.is_superuser: kwargs['queryset'] = Category.objects.all() else: kwargs['queryset'] = Category.objects.filter(editors=request.user) return db_field.formfield(**kwargs) return super(EntryAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) This way I can limit the categories to which a User can add Entry and it works perfect. Now the tricky part: On the Entry changelist/action page I want to show only those Entries which belong to current User's Category. I tried to do this using this method: def changelist_view(self, request, extra_context=None): if not request.user.is_superuser: self.queryset = self.queryset.filter(editors=request.user) But I get this error: AttributeError: 'function' object has no attribute 'filter' This is strange, because I thought it should be a typical QuerySet. Basically such methods are not well documented and digging through tons of Django code is not my favourite sport. Any ideas how can I achieve my goal?

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  • Can Django admin handle a one-to-many relationship via related_name?

    - by Mat
    The Django admin happily supports many-to-one and many-to-many relationships through an HTML <SELECT> form field, allowing selection of one or many options respectively. There's even a nice Javascript filter_horizontal widget to help. I'm trying to do the same from the one-to-many side through related_name. I don't see how it's much different from many-to-many as far as displaying it in the form is concerned, I just need a multi-select SELECT list. But I cannot simply add the related_name value to my ModelAdmin-derived field list. Does Django support one-to-many fields in this way? My Django model something like this (contrived to simplify the example): class Person(models.Model): ... manager = models.ForeignKey('self', related_name='staff', null=True, blank=True, ) From the Person admin page, I can easily get a <SELECT> list showing all possible staff to choose this person's manager from. I also want to display a multiple-selection <SELECT> list of all the manager's staff. I don't want to use inlines, as I don't want to edit the subordinates details; I do want to be able to add/remove people from the list. (I'm trying to use django-ajax-selects to replace the SELECT widget, but that's by-the-by.)

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  • Joomla 1.6 site cannot add a new extension through admin interface

    - by Ghlouw
    I'm having a very frustrating problem with my Joomla 1.6 site. I cannot add any new extensions through the admin interface. I have tried to upload the extension, or to use the search folder option or even the direct link. Neither of these options work, and all that happens is that the page tries to load forever until it finally timesout with a blank white page (no further error messages). I have tried this with multiple browsers (Chrome,FF,IE) and I have tried it with different extensions (modules, components, templates - all the same result). So I don't think it has anything to do with what I am uploading, but more likely the problem is something with the post action. I have also seen the exact same error occur when I try to update menu items or even create new menu items. I am not getting this error with a duplicate of the site in the dev environment, but only get this on my shared web hosting live server. This is on a Windows IIS / PHP / mySQL environment. Any help would be much appreciated!

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  • How to enable WordPress to have multiple sites without a re-direct

    - by user57039
    I'm using WordPress to manage my site and when the site does a re-direct, it slows down performance. For example, WordPress allows you a single default site, www.mycompany.com. If a user goes to mycompany.com, WP will re-direct it www.mycompany.com. Is there a way to configure WP so that it will listen on both www.mycompany.com and mycompany.com without redirects. The redirects are causing performance hits to the site.

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  • Is it possible to change the model name in the django admin site?

    - by luc
    Hello, I am translating a django app and I would like to translate also the homepage of the django admin site. On this page are listed the application names and the model class names. I would like to translate the model class name but I don't find how to give a user-friendly name for a model class. Does anybody know how to do that?

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  • Django admin breaking with non-default primary_key for model with a m2m relationship ?

    - by Gj
    I have a simple Post model with a m2m field to a Tag model. The Tag had for some reason to use a non default primary key. Inside the admin page for a Post, the labels for the multiple selection field for Tags appear, but not the input field itself. I also tried using the filter_horizontal for the tags, but still only the labels appear without the actual field. Any ideas why it breaks and/or workarounds? Thanks!

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  • Some OBI EE Tricks and Tips in the Admin Tool By Gerry Langton

    - by hamsun
    How to set the log level from a Session variable Initialization block As we know it is normal to set the log level non-zero for a particular user when we wish to debug problems. However sometimes it is inconvenient to go into each user’s properties in the Admin tool and update the log level. So I am showing a method which allows the log level to be set for all users via a session initialization block. This is particularly useful for anyone wanting an alternative way to set the log level. The screen shots shown are using the OBIEE 11g SampleApp demo but are applicable to any environment. Open the appropriate rpd in on-line mode and navigate to Manage Variables. Select Session Initialization Blocks, right click in the white space and create a New Initialization Block. I called the Initialization block Set_Loglevel . Now click on ‘Edit Data Source’ to enter the SQL. Chose the ‘Use OBI EE Server’ option for the SQL. This means that the SQL provided must use tables which have been defined in the Physical layer of the RPD, and whilst there is no need to provide a connection pool you must work in On-Line mode. The SQL can access any of the RPD tables and is purely used to return a value of 2. The ‘Test’ button confirms that the SQL is valid. Next, click on the ‘Edit Data Target’ button to add the LOGLEVEL variable to the initialization block. Check the ‘Enable any user to set the value’ option so that this will work for any user. Click OK and the following message will display as LOGLEVEL is a system session variable: Click ‘Yes’. Click ‘OK’ to save the Initialization block. Then check in the On-LIne changes. To test that LOGLEVEL has been set, log in to OBIEE using an administrative login (e.g. weblogic) and reload server metadata, either from the Analysis editor or from Administration > Reload Files and Metadata link. Run a query then navigate to Administration > Manage Sessions and click ‘View Log’ for the query just issued (which should be approximately the last in the list). A log file should exist and with LOGLEVEL set to 2 should include both logical and physical sql. If more diagnostic information is required then set LOGLEVEL to a higher value. If logging is required only for a particular analysis then an alternative method can be used directly from the Analysis editor. Edit the analysis for which debugging is required and click on the Advanced tab. Scroll down to the Advanced SQL clauses section and enter the following in the Prefix box: SET VARIABLE LOGLEVEL = 2; Click the ‘Apply SQL’ button. The SET VARIABLE statement will now prefix the Analysis’s logical SQL. So that any time this analysis is run it will produce a log. You can find information about training for Oracle BI EE products here or in the OU Learning Paths. Please send me an email at [email protected] if you have any further questions. About the Author: Gerry Langton started at Siebel Systems in 1999 working as a technical instructor teaching both Siebel application development and also Siebel Analytics (which subsequently became Oracle BI EE). From 2006 Gerry has worked as Senior Principal Instructor within Oracle University specialising in Oracle BI EE, Oracle BI Publisher and Oracle Data Warehouse development for BI.

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  • Admin Panel like Custom Framework

    - by bhuvin
    I want to Create a Framework , like Admin panel , which can rule almost all the aspects of what is shown on the frontend. For an (most basic) example: If suppose the links which are to be shown in a navigation area is passed from the server, with the order and the url , etc. The whole aim is to save the time on the tedious tasks. You can just start creating menus and start assigning pages to it. Give a url, actual files which are to be rendered (in case of static files.), in case of dynamic files, giving the file accordingly. And all this is fully server side manageable using different portlets, sort of things. So basic Roadmap is having : Areas like: Header Area - Which can contain logos, links etc. Navigation Area - Which can contains links and submenus. Content Area - Now this is where the tricky part is that that it has zones like: left, center & right. It contains Order in which it has to be displayed. So, when someday we want to change the way the articles appear on the page, we can do so easily, without any deployments. Now these zones can have n number of internal elements, like the word cloud, or the advertisement area. Footer Area: Again similar as Header Area. Currently there is a preexisting custom framework, which uses XSLT files for pulling out data from the server side. And it has the above capabilities. For example: If there's a grid it will be having a <table> tag embedded in the XSLT file. Now whatever might be the source of the data, we serialize this as XML and give it to the XSLT file and the html is derived from this and is appended to the layer in a page. The problem with this approach is: The XSLT conversion is occurring on the server side, so the server is responsible for getting the data, running XSLT transform, and append the html generated to the layer div. So, according to me, firstly this isn't the server's concern to do so. Secondly for larger applications this might be slower. Debugging isn't possible for XSLT transformation. So, whenever we face problems with data its always a bit of a trial & error method. Maintaining it is a bit of an eerie job i.e. styling changes, and other stuff. Adding dynamic values. Like JavaScript can't actually be very easily used in this. Secondly, we can't use JQuery or any other libraries with this since this is all occurring on the server. For now what I have thought about is using Templating - Javascript - JSON combination in place of XSLT, this will be offloaded to the client and the rendering will take place accordingly. This could solve the above problems and also could add mobile support for the same. Only problem which I could think of is that: It is much work and adding new portlets on the go needs to be looked into. What could be the alternatives for this? What kind of problems are there with the JavaScript approach? What are the different ways to implement the same? Are there any existing frameworks for similar usage?

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  • UnicodeEncodeError when uploading files in Django admin

    - by Samuel Linde
    Note: I asked this question on StackOverflow, but I realize this might be a more proper place to ask this kind of question. I'm trying to upload a file called 'Testaråäö.txt' via the Django admin app. I'm running Django 1.3.1 with Gunicorn 0.13.4 and Nginx 0.7.6.7 on a Debian 6 server. Database is PostgreSQL 8.4.9. Other Unicode data is saved to the database with no problem, so I guess the problem must be with the filesystem somehow. I've set http { charset utf-8; } in my nginx.conf. LC_ALL and LANG is set to 'sv_SE.UTF-8'. Running 'locale' verifies this. I even tried setting LC_ALL and LANG in my nginx init script just to make sure locale is set properly. Here's the traceback: Traceback (most recent call last): File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/handlers/base.py", line 111, in get_response response = callback(request, *callback_args, **callback_kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 307, in wrapper return self.admin_site.admin_view(view)(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 93, in _wrapped_view response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/views/decorators/cache.py", line 79, in _wrapped_view_func response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 197, in inner return view(request, *args, **kwargs) File "/srv/django/letebo/app/cms/admin.py", line 81, in change_view return super(PageAdmin, self).change_view(request, obj_id) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 28, in _wrapper return bound_func(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 93, in _wrapped_view response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 24, in bound_func return func(self, *args2, **kwargs2) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/transaction.py", line 217, in inner res = func(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 985, in change_view self.save_formset(request, form, formset, change=True) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 677, in save_formset formset.save() File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 482, in save return self.save_existing_objects(commit) + self.save_new_objects(commit) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 613, in save_new_objects self.new_objects.append(self.save_new(form, commit=commit)) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 717, in save_new obj.save() File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 460, in save self.save_base(using=using, force_insert=force_insert, force_update=force_update) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 504, in save_base self.save_base(cls=parent, origin=org, using=using) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 543, in save_base for f in meta.local_fields if not isinstance(f, AutoField)] File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/fields/files.py", line 255, in pre_save file.save(file.name, file, save=False) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/fields/files.py", line 92, in save self.name = self.storage.save(name, content) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 48, in save name = self.get_available_name(name) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 74, in get_available_name while self.exists(name): File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 218, in exists return os.path.exists(self.path(name)) File "/srv/.virtualenvs/letebo/lib/python2.6/genericpath.py", line 18, in exists st = os.stat(path) UnicodeEncodeError: 'ascii' codec can't encode characters in position 52-54: ordinal not in range(128) I tried running Gunicorn with debugging turned on, and the file uploads without any problem at all. I suppose this must mean that the issue is with Nginx. Still beats me where to look, though. Here are the raw response headers from Gunicorn and Nginx, if it makes any sense: Gunicorn: HTTP/1.1 302 FOUND Server: gunicorn/0.13.4 Date: Thu, 09 Feb 2012 14:50:27 GMT Connection: close Transfer-Encoding: chunked Expires: Thu, 09 Feb 2012 14:50:27 GMT Vary: Cookie Last-Modified: Thu, 09 Feb 2012 14:50:27 GMT Location: http://my-server.se:8000/admin/cms/page/15/ Cache-Control: max-age=0 Content-Type: text/html; charset=utf-8 Set-Cookie: messages="yada yada yada"; Path=/ Nginx: HTTP/1.1 500 INTERNAL SERVER ERROR Server: nginx/0.7.67 Date: Thu, 09 Feb 2012 14:50:57 GMT Content-Type: text/html; charset=utf-8 Transfer-Encoding: chunked Connection: close Vary: Cookie 500 UPDATE: Both locale.getpreferredencoding() and sys.getfilesystemencoding() outputs 'UTF-8'. locale.getdefaultlocale() outputs ('sv_SE', 'UTF8'). This seem correct to me, so I'm still not sure why I keep getting these errors.

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  • Adding links to full change forms for inline items in django admin?

    - by David Eyk
    I have a standard admin change form for an object, with the usual StackedInline forms for a ForeignKey relationship. I would like to be able to link each inline item to its corresponding full-sized change form, as the inline item has inlined items of its own, and I can't nest them. I've tried everything from custom widgets to custom templates, and can't make anything work. So far, the "solutions" I've seen in the form of snippets just plain don't seem to work for inlines. I'm getting ready to try some DOM hacking with jQuery just to get it working and move on. I hope I must be missing something very simple, as this seems like such a simple task! Using Django 1.2.

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  • Creating an Admin directory in Rails

    - by matsko
    I've been developing the CMS backend for a website for a few weeks now. The idea is to craft everything in the backend first so that it can manage the database and information that will be displayed on the main website. As of now, I currently have all my code setup in the normal rails MVC structure. So the users admin is /users and videos is /videos. My plans are to take the code for this and move it to a /admin directory. So the two controllers above would need to be accessed by /admin/users and /admin/videos. I'm not sure how todo the ruote (adding the /admin as a prefix) nor am I sure about how to manage the logic. What I'm thinking of doing is setting up an additional 'middle' controller that somehow gets nested between the ApplicationControler and the targetted controller when the /admin directory is accessed. This way, any additional flags and overloaded methods can be spawned for the /admin section only (I believe I could use a filter too for this). If that were to work, then the next issue would be separating the views logic (but that would just be renaming folders and so on). Either I do it that way or I have two rails instances that share the MVC code between them (and I guess the database too), but I fear that would cause lots of duplication errors. Any ideas as to how I should go about doing this? Many thanks!

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  • Alternate User select interface in django admin to reduce page size on large site?

    - by David Eyk
    I have a Django-based site with roughly 300,000 User objects. Admin pages for objects with a ForeignKey field to User take a very long time to load as the resulting form is about 6MB in size. Of course, the resulting dropdown isn't particularly useful, either. Are there any off-the-shelf replacements for handling this case? I've been googling for a snippet or a blog entry, but haven't found anything yet. I'd like to have a smaller download size and a more usable interface.

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  • why isn't my id showing up in django admin list?

    - by FurtiveFelon
    Hi all, I have a class Task(models.Model), and i didn't define id field explicitly (since it defines automatically for you). I checked in the database, it exists for the Task. Now i would like to display it in the list via list_display property in admin.ModelAdmin. I have a bunch of things in there, only id is not showing up for any of the rows i have. Everything else works fine. Anyone know anything special i have to do to get id to display? Thanks a lot! Jason

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  • What's the straightforward way to implement one to many editing in list_editable in django admin?

    - by Nate Pinchot
    Given the following models: class Store(models.Model): name = models.CharField(max_length=150) class ItemGroup(models.Model): group = models.CharField(max_length=100) code = models.CharField(max_length=20) class ItemType(models.Model): store = models.ForeignKey(Store, on_delete=models.CASCADE, related_name="item_types") item_group = models.ForeignKey(ItemGroup) type = models.CharField(max_length=100) Inline's handle adding multiple item_types to a Store nicely when viewing a single Store. The content admin team would like to be able to edit stores and their types in bulk. Is there a simple way to implement Store.item_types in list_editable which also allows adding new records, similar to horizontal_filter? If not, is there a straightforward guide that shows how to implement a custom list_editable template? I've been Googling but haven't been able to come up with anything. Also, if there is a simpler or better way to set up these models that would make this easier to implement, feel free to comment.

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  • How to show raw_id value of a ManyToMany relation in the Django admin?

    - by luc
    Hello, I have an app using raw_id on both ForeignKeyField and ManyToManyField. The admin displays the value of the foreign key on the right of the edit box. Unfortunatey, it doesn't work with ManyToMany. I've checked the code and I think that it is the normal behavior. However I would like to know if someone has an easy tip to change this behavior? Thanks in advance Update: I've tried to subclass the ManyToManyRawIdWidget but I don't know how to say that the raw_id_fields should use my custom widget. formfield_overrides doesn't seem to work with raw_id fields

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  • Apache Solr Admin on Tomcat Deployed in WebApps Directory

    - by KM01
    I am trying to get Apache Solr to work on Redhat6 and Tomcat6 (using these instructions), but get this error when browsing to the admin section, http://localhost:8080/solr-example/admin: HTTP Status 404 - missing core name in path type Status report message missing core name in path description The requested resource (missing core name in path) is not available. http://localhost:8080/solr-example loads fine, with a link to "Solr Admin." My setup is as follows: tomcat6: /etc/tomcat6 Solr: /app/solr/example I have a solr-example.xml in /etc/tomcat6/Catalina/localhost/, which reads: <?xml version="1.0" encoding="utf-8"?> <Context docBase="/app/solr/example/apache-solr-3.4.0.war" debug="0" crossContext="true"> <Environment name="solr/home" type="java.lang.String" value="/app/solr/example" override="true"/> </Context> I don't see anything in the logs (/var/log/tomcat6) ... only entires in catalina.out are regarding the starting and stopping of tomcat6. My questions are: 1.What else do I need to do to get "Solr Admin" to work under Tomcat? 2.Where are these "cores" supposed to be specified? I see an entry in /app/solr/example/solr/solr.xml ? <solr persistent="false"> adminPath: RequestHandler path to manage cores. If 'null' (or absent), cores will not be manageable via request handler <cores adminPath="/admin/cores" defaultCoreName="collection1"> <core name="collection1" instanceDir="." /> </cores> </solr> 3.How do I got about ensuring that logs are working correctly? I can't find logs that contain mention of the 404 above. Update in response to @quanta's comment: Downloaded former (apache-solr-3.4.0.tgz) dataDir was not set, now set to: <dataDir>${solr.data.dir:../solr/data}</dataDir> JAVA_OPTS: /usr/lib/jvm/java/bin/java -classpath :/usr/share/tomcat6/bin/bootstrap.jar:/usr/share/tomcat6/bin/tomcat-juli.jar:/usr/share/java/commons-daemon.jar -Dcatalina.base=/usr/share/tomcat6 -Dcatalina.home=/usr/share/tomcat6 -Djava.endorsed.dirs= -Djava.io.tmpdir=/var/cache/tomcat6/temp -Djava.util.logging.config.file=/usr/share/tomcat6/conf/logging.properties -Djava.util.logging.manager=org.apache.juli.ClassLoaderLogManager org.apache.catalina.startup.Bootstrap start catalina.out contains no indication of the above error

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  • Best way to re-use the same django models and admin for multiple apps

    - by kepioo
    Given a reference app ( called guide), how can I create additional apps that will reuse the same model/admin/views than guide - the motivation behind is to be able to individually control each subapp. guide guideApp1 exact same models/admin/views than guide guideApp2 exact same models/admin/views than guide in the Admin site, I should have : 1 section for guideApp1 with all the tables defined in guide, that applies to guideApp1 1 section for guideApp12 with all the tables defined in guide, that applies to guideApp2

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  • after registration admin will approve in oscommerece

    - by manish
    Hi... I want the solution for Oscommerece.after registration ,user will wait for admin approval. when user will register then one mail go into the admin mail by which admin can know there is one request for approval. By default in oscommerece there is no facility to admin approval. Please help me as soon as possible Thanks Manish

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