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• Changes in licence in forked project what are my rights?

  - by Wes

  Hi I'm intrested in using the apparently now defunct app-mdi libray in a flex application for a paying customer. http://sourceforge.net/projects/appmdi/ It appears that the app-mdi project has been forked from flex-mdi and indeed the code has so much in common it would appear almost identical to the origional code. Now in the original source flex-mdi the following licence appears in the source code /* Copyright (c) 2007 FlexMDI Contributors. See: http://code.google.com/p/flexmdi/wiki/ProjectContributors Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE. */ However in the app-mdi library on the same file the following licence appears. Copyright (c) 2010, TRUEAGILE All rights reserved. Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met: Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer. Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution. Neither the name of the TRUEAGILE nor the names of its contributors may be used to endorse or promote products derived from this software without specific prior written permission. THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. */ Now I've no problem with the licence except for the line. Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution. The copyright notice in its entireity makes no sense in binary material. Specifically talking about redistobutions in the binary form. Finally the question is what exactly has to be shown on web clients who access softare that utilises this library? Also is changing the licence in this manner actually allowed?

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• Extended Django base-class with multiple instances

  - by Gijs

  I'm modeling a simple movie database using Django. models.py defines a base model Person. I extend Person into Actor and Director, which works as I imagined. Persons must be unique. When (in the Admin) I create an instance of Actor, and this person is also a Director, it won't save because of the unique = True. Any ideas how to solve this problem? (generic foreign keys?) Thx

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• How can I limit ModelAdmin queryset to user-editable items when using object permissions?

  - by brookstravis

  I'm trying to implement Florian's object permissions system from his article on Django Advent, but I'm running into an issue trying to limit the queryset returned to only items the user has permission to edit/view. Florian mentions it in his section on wrapping the admin, but skips over it. I can't see a good way to filter the queryset. Am I missing something?

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• Making only a part of model field available in Django

  - by Hellnar

  Hello I have a such model: GENDER_CHOICES = ( ('M', 'Male'), ('F', 'Female') ) class Profile(models.Model): user = models.ForeignKey(User) gender = models.CharField(max_length=1, choices=GENDER_CHOICES) class FrontPage(models.Model): female = models.ForeignKey(User,related_name="female") male = models.ForeignKey(User,related_name="male") Once I attempt to add a new FrontPage object via the Admin page, I can select "Female" profiles for the male field of FrontPage, how can I restrict that? Thanks

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• Changing models in django results in broken database?

  - by Rhubarb

  I have added and removed fields in my models.py file and then run manage.py syncdb. Usually I have to quit out of the shell and restart it before syncdb does anything. And then even after that, I am getting errors when trying to access the admin pages, it seems that certain new fields that I've added still don't show up in the model: Caught an exception while rendering: no such column: mySite_book.Title

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• extending satchmo user profile

  - by z3a

  I'm trying to extend the basic user registration form and profile included in satchmo store, but I'm in problems with that. This what I've done: Create a new app "extendedprofile" Wrote a models.py that extends the satchmo_store.contact.models class and add the custom name fields. wrote an admin.py that unregister the Contact class and register my newapp but this still showing me the default user profile form. Maybe some one can show me the correct way to do this?

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• django internationalisation

  - by ha22109

  Hello , I am doing django admin internationalization .I am able to do it perfectly.But my concern is that in the address bar it is showing the app label in tranlated form which is not in us acii .Is this the problem with django or i m doing something wrong.

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• django internationalisation loading error

  - by ha22109

  Hello All, I m not able to change the locale of django -admin when i switch to different locale from browser.But if i mention in my settings.py the language code then it is working but browser locale doesnot have any impact on it #setting.py LANGUAGES = ( ('ar', gettext_noop('Arabic')), ('ja', gettext_noop('japanese')), ('bg', gettext_noop('Bulgarian')), LANGUAGE_CODE = ''# LANGUAGE_COOKIE_NAME = 'django_language' LOCALE_PATHS = ()

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• Django save method

  - by Marijus

  So I have a model with a FileField for excel spreadsheet. What I need to do this add another column in this spreadsheet, in each row let user pick from a drop-down list then save it and display it in html. All the picking and uploading will happen through the admin interface. So I have figured out way how to display a spreadsheet in html, however I have no idea how to write this save method. I could really use some hints and tips..

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• how to limit the foreignkey dropdown with constraints?

  - by FurtiveFelon

  Hi all, I have a database which keeps track of interaction between two different teams (represented in the admin interface by two different groups). For some fields, i have a foreignkey to Users database, and i would like to limit the dropdown people to only the specific groups. If anyone have any suggestions, it would be much appreciated! Jason

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• Any way to add tabbed forms in django administration site ?

  - by tomjerry

  When using Django "out-of-the-box" administration forms, the "change form" pages can be rather long for complex models (with a lot of fields). I would like to use tabs in the "change form", so things can be more readable (group fields by tabs...) Instead of doing it all by myself, by modifiying the 'change_form.html' admin template, I was wondering whether somebody has already done that and would like to share the code, or whether an existing Django-plugin already exist. Thanks in advance for you answer

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• Unlock device, display a text, then lock again

  - by Waza_Be

  For the need of my application, I need to display a message on the screen even if the lockscreen is enabled, then wait 3 seconds, than I have to lock again the phone as I don't want it to make unwanted phone calls in your pockets. First part is easy: if (PreferenceManager.getDefaultSharedPreferences( getBaseContext()).getBoolean("wake", false)) { KeyguardManager kgm = (KeyguardManager) getSystemService(Context.KEYGUARD_SERVICE); boolean isKeyguardUp = kgm.inKeyguardRestrictedInputMode(); WakeLocker.acquire(ProtoBenService.this); Intent myIntent = new Intent(ProtoBenService.this,LockActivity.class); myIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK); if (isKeyguardUp) { ProtoBenService.this.startActivity(myIntent); } else Toast.makeText(ProtoBenService.this.getBaseContext(), intention, Toast.LENGTH_LONG).show(); WakeLocker.release(); } With this class: public abstract class WakeLocker { private static PowerManager.WakeLock wakeLock; public static void acquire(Context ctx) { if (wakeLock != null) wakeLock.release(); PowerManager pm = (PowerManager) ctx.getSystemService(Context.POWER_SERVICE); wakeLock = pm.newWakeLock(PowerManager.FULL_WAKE_LOCK | PowerManager.ACQUIRE_CAUSES_WAKEUP | PowerManager.ON_AFTER_RELEASE, "CobeIm"); wakeLock.acquire(); } public static void release() { if (wakeLock != null) wakeLock.release(); wakeLock = null; } } And the Activity: public class LockActivity extends Activity { @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); Window window = getWindow(); window.addFlags(WindowManager.LayoutParams.FLAG_DISMISS_KEYGUARD); window.addFlags(WindowManager.LayoutParams.FLAG_TURN_SCREEN_ON); window.addFlags(WindowManager.LayoutParams.FLAG_KEEP_SCREEN_ON); TextView tv = new TextView(this); tv.setText("This is working!"); tv.setTextSize(45); setContentView(tv); Runnable mRunnable; Handler mHandler = new Handler(); mRunnable = new Runnable() { @Override public void run() { LockActivity.this.finish(); } }; mHandler.postDelayed(mRunnable, 3 * 1000); } } So, this is nice, the phone can display my text! The only problem comes when I want to lock again the phone, it seems that locking the phone is protected by the system... Programmatically turning off the screen and locking the phone how to lock the android programatically I think that my users won't understand the Device Admin and won't be able to activate it. Is there any workaround to lock the screen without the Device Admin stuff?

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• Django menu item sorting

  - by doktorno

  Hi i've got MenuItem model : MenuItem(models.Model) name=models.CharField(max_length=50) url = models.URLField() position = models.IntegerField() Class Meta: ordering =['position'] then i'm retriving it by MenuItem.objects.all() My question is how can i make any user friendly interface in admin panel to allow sorting MenuItems - for example list with + and - buttons to move MenuItem up and down ....

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• Permission to view, but not to change! - Django

  - by RadiantHex

  Hi folks, is it possible to give users the permission to view, but not to change or delete. currently in the only permissions I see are "add", "change" and "delete"... but there is no "read/view" in there. I really need this as some users will only be able to consult the admin panel, in order to see what has been added in. Help would be amazing!

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• Web application creation in IIS7 via MS.Web.Admin

  - by Jon Ownbey

  I am attempting to create seperate workflow instances as applications in IIS7 using the Microsoft.Web.Administration dll. When it attempts to add the Application to the Site ApplicationsCollection I get a COM error: "Invalid application path\r\n" using (ServerManager manager = new ServerManager()) { var site = manager.Sites.Where(x => x.Name == Properties.Settings.Default.WorkflowWebsiteName).Single(); StringBuilder stringBuilder = new StringBuilder() .Append(m_workflowDefinition.AccountId) .Append("/") .Append(m_workflowDefinition.WorkflowDefinitionId) .Append("/") .Append(m_workflowDefinition.Version) .Append("/"); string virtualPath = stringBuilder.ToString(); string physicalPath = Properties.Settings.Default.ApplicationPoolString + virtualPath.Replace("/", "\\"); if (!Directory.Exists(physicalPath)) Directory.CreateDirectory(physicalPath); //Create the workflow service definition file using (StreamWriter writer = new StreamWriter(Path.Combine(physicalPath, m_workflowDefinition.WorkflowName + WORKFLOW_FILE_EXTENSION))) { writer.Write(m_workflowDefinition.Definition); } //Copy dependencies string dependencyPath = m_workflowDefinition.DependenciesPath; CopyAll(new DirectoryInfo(dependencyPath), new DirectoryInfo(physicalPath)); //Create a new IIS application for the workflow var apps = site.Applications.Where(x => x.Path == virtualPath); if (apps.Count() > 0) { site.Applications.Remove(apps.Single()); } Application app = site.Applications.Add(virtualPath, physicalPath); app.ApplicationPoolName = "Workflow AppPool"; app.EnabledProtocols = PROTOCOLS; manager.CommitChanges(); } The value assigned to virtualPath is like: "something/something/something" and for physicalPath it is "c:\inetpub\wwwroot\Workflow\something\something\something". Any ideas? Any help is greatly appreciated.

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• proxy pass for activeMQ

  - by user1172482

  I have a apache server that I'm trying to use for proxy access my activeMQ admin page. I am able to load the inital landing page properly, but I can't seem to load any of the sub-pages (Queues, Connections, etc.). My proxypass rules on the apache server are the following: ProxyPass /foo http://10.5.124.108:8161/admin ProxyPassReverse /foo http://10.5.124.108:8161/admin The activeMQ installation included a activemq-httpd.conf file in /etc/httpd/conf.d/. Proxy connections there are enabled: ProxyRequests On ProxyVia On <Proxy *> Allow from all Order allow,deny </Proxy> ProxyPass /admin http://localhost:8161/admin ProxyPassReverse /admin http://localhost:8161/admin ProxyPass /message http://localhost:8161/admin/send ProxyPassReverse /message http://localhost:8161/admin/send From what I've read the proxypass rules should be recursive (the rule for /foo should also work for /foo/bar). Is there something else that I'm missing here that's preventing me from accessing pages beyond the initial admin landing page?

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• Automatically create an admin user when running Django's ./manage.py syncdb

  - by a paid nerd

  My project is in early development. I frequently delete the database and run manage.py syncdb to set up my app from scratch. Unfortunately, this always pops up: You just installed Django's auth system, which means you don't have any superusers defined. Would you like to create one now? (yes/no): Then you have supply a username, valid email adress and password. This is tedious. I'm getting tired of typing test\[email protected]\ntest\ntest\n. How can I automatically skip this step and create a user programatically when running manage.py syncdb ?

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• Help with understanding generic relations in Django (and usage in Admin)

  - by saturdayplace

  I'm building a CMS for my company's website (I've looked at the existing Django solutions and want something that's much slimmer/simpler, and that handles our situation specifically.. Plus, I'd like to learn this stuff better). I'm having trouble wrapping my head around generic relations. I have a Page model, a SoftwareModule model, and some other models that define content on our website, each with their get_absolute_url() defined. I'd like for my users to be able to assign any Page instance a list of objects, of any type, including other page instances. This list will become that Page instance's sub-menu. I've tried the following: class Page(models.Model): body = models.TextField() links = generic.GenericRelation("LinkedItem") @models.permalink def get_absolute_url(self): # returns the right URL class LinkedItem(models.Model): content_type = models.ForeignKey(ContentType) object_id = models.PositiveIntegerField() content_object = generic.GenericForeignKey('content_type', 'object_id') title = models.CharField(max_length=100) def __unicode__(self): return self.title class SoftwareModule(models.Model): name = models.CharField(max_length=100) description = models.TextField() def __unicode__(self): return self.name @models.permalink def get_absolute_url(self): # returns the right URL This gets me a generic relation with an API to do page_instance.links.all(). We're on our way. What I'm not sure how to pull off, is on the page instance's change form, how to create the relationship between that page, and any other extant object in the database. My desired end result: to render the following in a template: <ul> {% for link in page.links.all %} <li><a href='{{ link.content_object.get_absolute_url() }}'>{{ link.title }}</a></li> {% endfor%} </ul> Obviously, there's something I'm unaware of or mis-understanding, but I feel like I'm, treading into that area where I don't know what I don't know. What am I missing?

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• django: Display image in admin interface

  - by Oleg Tarasenko

  Hi, I've defined a model which contains a link an image. Is there a way to display the image in the model items list. (e.g. if I defined an article this way: class Article(models.Model): url = models.CharField(max_length = 200, unique = True, help_text="/lessons/") title = models.CharField(max_length = 500) img = models.CharField(max_length = 100) # Contains path to image def __unicode__(self): return u"%s" %title ) Is there a way to display image together with title?

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• Authlogic Current User Question - hiding admin links...

  - by bgadoci

  I think I am missing something while using the Authlogic gem w/ Rails. To set the stage I have multiple users and each user can create posts and comments. Upon the display of a post or comment I would like to give the user who created them the option to edit or destroy. I am successfully using the following code to hide and show elements based on if a user is logged in or not but can't seem to find out how to only show these links to the actual user who created them...not any user that is logged in. <% if current_user %> <%= link_to 'Edit', edit_question_path(question) %> | <%= link_to 'Destroy', question, :confirm => 'Are you sure?', :method => :delete %> <% else %> <p>nothing to see here</p> <% end %> Here is the def of current_user located in the application controller in case I need to change something here. class ApplicationController < ActionController::Base helper :all # include all helpers, all the time protect_from_forgery # See ActionController::RequestForgeryProtection for details# helper_method :current_user private def current_user_session return @current_user_session if defined?(@current_user_session) @current_user_session = UserSession.find end def current_user return @current_user if defined?(@current_user) @current_user = current_user_session && current_user_session.record end end

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• Can IIS admin change password of Windows Service account

  - by sweta Jha

  We have a service account defined for anonymous access which is used for several web sites hosted on the web server. This account has access to several network resources like report server, file servers and so on. While deploying a new web site, we used the same service account for anonymous access. IIS takes the username/password for the account and then a dialog opens for confirm password. Accidently, we gave a wrong password in both the text boxes, the new site with wrong password is working fine but all other previously hosted sites which were using the service account, started giving the unautorized access error. Is it possible that when we entered wrong password for the new web site, the password of the account got reset and all sites stopped functioning?

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• Notify Admin of an update in Sharepoint?

  - by DotnetDude

  Is there a way to get an update by email when a user updates a wiki page or adds items into the document library?

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• Set up proxy for vpn server on ubuntu server 12.4

  - by Morteza Soltanabadiyan

  I have a vpn server with HTTPS, L2TP, OPENVPN, and PPTP. I want to set up a proxy on the server, so all connection that comes from vpn clients, they will use that. I created the following bash script file for it, but the proxy isn't working. gsettings set org.gnome.system.proxy mode 'manual' gsettings set org.gnome.system.proxy.http enabled true gsettings set org.gnome.system.proxy.http host 'cproxy.anadolu.edu.tr' gsettings set org.gnome.system.proxy.http port 8080 gsettings set org.gnome.system.proxy.http authentication-user 'admin' gsettings set org.gnome.system.proxy.http authentication-password 'admin' gsettings set org.gnome.system.proxy use-same-proxy true export http_proxy=http://admin:[email protected]:8080 export https_proxy=http://admin:[email protected]:8080 export HTTP_PROXY=http://admin:[email protected]:8080 export HTTPS_PROXY=http://admin:[email protected]:8080 What to do to make a global proxy for server and all vpn clients to use it automatically?

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• How to solve 404 for static files with Django and Nginx?

  - by Lucio

  I setup a Trusty VM with Django + Nginx (other stuffs too). The server is working, I get the "Welcome to Django" page. But when I enter to servername/admin it loads the HTML page but fails to load the static content. And the admin page have this links to static content: <link rel="stylesheet" type="text/css" href="/static/admin/css/base.css" /> <link rel="stylesheet" type="text/css" href="/static/admin/css/login.css" /> Both of the CSS files give me 404, as the Nginx log shows: 192.168.56.1 - - [05/Jun/2014:12:04:09 -0300] "GET /admin HTTP/1.1" 301 5 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:29.0) Gecko/20100101 Firefox/29.0" 192.168.56.1 - - [05/Jun/2014:12:04:09 -0300] "GET /admin/ HTTP/1.1" 200 833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:29.0) Gecko/20100101 Firefox/29.0" 192.168.56.1 - - [05/Jun/2014:12:04:10 -0300] "GET /static/admin/css/base.css HTTP/1.1" 404 142 "http://ubuntu-server/admin/" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:29.0) Gecko/20100101 Firefox/29.0" 192.168.56.1 - - [05/Jun/2014:12:04:10 -0300] "GET /static/admin/css/login.css HTTP/1.1" 404 142 "http://ubuntu-server/admin/" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:29.0) Gecko/20100101 Firefox/29.0" I think that the error is on my nginx.conf file, but do not know how to solve it.

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• set proxy for vpn server on ubuntu server 12.4

  - by Morteza Soltanabadiyan

  I have a vpn server with HTTPS, L2TP , OPENVPN , PPTP. i want to set proxy in the server so all connection that comes from vpn clients use the proxy that i set in my server. I made a bash script file for it , but proxy not working. gsettings set org.gnome.system.proxy mode 'manual' gsettings set org.gnome.system.proxy.http enabled true gsettings set org.gnome.system.proxy.http host 'cproxy.anadolu.edu.tr' gsettings set org.gnome.system.proxy.http port 8080 gsettings set org.gnome.system.proxy.http authentication-user 'admin' gsettings set org.gnome.system.proxy.http authentication-password 'admin' gsettings set org.gnome.system.proxy use-same-proxy true export http_proxy=http://admin:[email protected]:8080 export https_proxy=http://admin:[email protected]:8080 export HTTP_PROXY=http://admin:[email protected]:8080 export HTTPS_PROXY=http://admin:[email protected]:8080 Now , i dont know what to do to make a global proxy for server and all vpn clients use it automatically.

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