If you want to be able to specify ManyToMany relation without making it required just use `blank=True`:

    class Group(models.Model):
        ...
        events = models.ManyToManyField(Event, blank=True)

I&#39;m running a curl request on an eXist database through php. The dataset is very large, and as a result, the database consistently takes a long amount of time to return an XML response. To fix that, we set up a curl request, with what is supposed to be a long timeout.

    $ch = curl_init();
    $headers[&quot;Content-Length&quot;] = strlen($postString);
    $headers[&quot;User-Agent&quot;] = &quot;Curl/1.0&quot;;

    curl_setopt($ch, CURLOPT_URL, $requestUrl);
    curl_setopt($ch, CURLOPT_HEADER, false);
    curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_USERPWD, &#39;admin:&#39;);
    curl_setopt($ch,CURLOPT_TIMEOUT,1000);
    $response = curl_exec($ch);
    curl_close($ch);
However, the curl request consistently ends before the request is completed (&lt;1000 when requested via a browser). Does anyone know if this is the proper way to set timeouts in curl?

Setting Curl&#39;s Timeout in PHP

I was wondering if someone could help me out. I am trying to create a custom AlertDialog. In order to do this, I added the following line of code in styles.xml

    &lt;resources&gt;
     &lt;style name=&quot;CustomAlertDialog&quot; parent=&quot;android:Theme.Dialog.Alert&quot;&gt;
      &lt;item name=&quot;android:windowBackground&quot;&gt;@drawable/color_panel_background&lt;/item&gt;
     &lt;/style&gt;
    &lt;/resources&gt;

* color_panel_background.9.png is located in drawable folder. This is also available in Android SDK res folder.

The following is the main activity.

    package com.customdialog;

    import android.app.Activity;
    import android.app.AlertDialog;
    import android.app.Dialog;
    import android.content.DialogInterface;
    import android.os.Bundle;

    public class CustomDialog extends Activity {
        /** Called when the activity is first created. */
        @Override
        public void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.main);
        
            this.setTheme(R.style.CustomAlertDialog);
            AlertDialog.Builder builder = new AlertDialog.Builder(this);
            builder.setMessage(&quot;HELLO!&quot;);
            builder .setCancelable(false)
              .setPositiveButton(&quot;Yes&quot;, new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                   //MyActivity.this.finish();
               }
           })
           .setNegativeButton(&quot;No&quot;, new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                   //dialog.cancel();
               }
           });

            AlertDialog alertdialog = builder.create();
            alertdialog.show();
        }
    }

In order to apply the theme to an AlertDialog, I had to set the theme to the current context.

However, I just can&#39;t seem to get the app to show customized AlertDialog. Can anyone help me out with this?

How to change theme for AlertDialog

When you have a many-to-many relationship (`related_name`, not `through`) and you are trying to use the admin interface you are required to enter one of the relationships even though it does not have to exist for you to create the first entry.

I&#39;m creating an app that is an event organizer. Imagine we had `Event` and `Group` models, bound with many-to-many relationship.

Django [`related_name`][1] creates another table with the indices of the two other tables.   
**But I see no reason why this extra table *has* to be populated.**

If I work with the database through phpMyAdmin I can create a `Group` without registering an `Event`, since the connection between the two is only through a separate table, and there is no database value enforcement at given level.

How do I make the admin interface this realize it?  
**How do I make the many-to-many field optional in Django?**


  [1]: https://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.ForeignKey.related_name

How do I make many-to-many field optional in Django?

When you have a many-to-many relationship (<code>related_name</code>, not <code>through</code>) and you are trying to use the admin interface you are required to enter one of the relationships even though it does not have to exist for you to create the first entry.
I'm creating an app that is an event organizer. Imagine we had <code>Event</code> and <code>Group</code> models, bound with many-to-many relationship.
Django <a href="https://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.ForeignKey.related_name" target="_blank" rel="noopener noreferrer"><code>related_name</code></a> creates another table with the indices of the two other tables. 
But I see no reason why this extra table has to be populated.
If I work with the database through phpMyAdmin I can create a <code>Group</code> without registering an <code>Event</code>, since the connection between the two is only through a separate table, and there is no database value enforcement at given level.
How do I make the admin interface this realize it? 
How do I make the many-to-many field optional in Django?

I want to get an object from the database if it already exists (based on provided parameters) or create it if it does not.

Django&#39;s [`get_or_create`](https://docs.djangoproject.com/en/1.10/ref/models/querysets/#get-or-create) (or [source](https://github.com/django/django/blob/master/django/db/models/query.py#L462)) does this.  Is there an equivalent shortcut in SQLAlchemy?

I&#39;m currently writing it out explicitly like this:

    def get_or_create_instrument(session, serial_number):
        instrument = session.query(Instrument).filter_by(serial_number=serial_number).first()
        if instrument:
            return instrument
        else:
            instrument = Instrument(serial_number)
            session.add(instrument)
            return instrument

Does SQLAlchemy have an equivalent of Django&#39;s get_or_create?

If I set a session variable in Django, like: 

    request.session[&quot;name&quot;] = &quot;name&quot;

Is there a way I can access it from within a template, or do I have to retrieve it from within a view, and then pass it to a template?

Asking because I have around 10 little session variables that I&#39;d like to access within a template, and passing all 10 from the view to the template could get a bit messy. 

(I have to use session variables because it&#39;s a HttpResponseRedirect, but storing the variables in a database is overkill for my purposes.)

So - any way to grab session variables directly within a template?


Django: accessing session variables from within a template?

I am curious which one would be better fitting as a currency field ? I will do simple operations such as taking difference, the percentage between old and new prices. I plan to keep two digits after the zero (ie 10.50) and majority of the time if these digits are zero, I will be hiding these numbers and display it as &quot;10&quot;

ps: Currency is NOT dollar based :)

Django: FloatField or DecimalField for Currency?

How do I change some models name from &quot;Categorys&quot; to &quot;Categories&quot; on admin site in the new dev django version?
In the old version (whithout admin sites and admin models) you could just do this;
http://www.the-dig.com/blog/post/customize-plural-name-django-admin/

However - now setting verbose_name_plural inside my modeladmin based class does nothing.
Anyone encouter the same issue?

Django fix Admin plural

This is my model:

    class Position(models.Model):
        map = models.ForeignKey(Map,primary_key=True)
        #members=models.CharField(max_length=200)
        LatLng = models.CharField(max_length=40000)
        infowindow = models.CharField(max_length=40000)


But it can&#39;t run. What is the max size of the `max_length` parameter?

What is the max size of &#39;max_length&#39; in Django?

In Django&#39;s `ModelAdmin`, I need to display forms customized according to the permissions an user has. Is there a way of getting the current user object into the form class, so that i can customize the form in its `__init__` method?

I think saving the current request in a thread local would be a possibility but this would be my last resort because I&#39;m thinking it is a bad design approach.

Django: How to get current user in admin forms?

I have ceated several django apps and stuffs for my own fund and so far everything has been working fine. 

Now I just created new project (django 1.2.1) and have run into trouble from 1st moments. 

I created new app - game and new model Game. I created admin.py and put related stuff into it. Ran syncdb and went to check into admin. Model did not show up there. 

I proceeded to check and doublecheck and read through previous similar threads:
https://stackoverflow.com/questions/1839927/registered-models-do-not-show-up-in-admin
https://stackoverflow.com/questions/1694259/django-app-not-showing-up-in-admin-interface

But as far as I can tell, they don&#39;t help me either. Perhaps someone else can point this out for me. 

models.py in game app:

    # -*- coding: utf-8 -*-
    from django.db import models
    
    class Game(models.Model):
          type = models.IntegerField(blank=False, null=False, default=1)
          teamone = models.CharField(max_length=100, blank=False, null=False)
          teamtwo = models.CharField(max_length=100, blank=False, null=False)
          gametime = models.DateTimeField(blank=False, null=False)

admin.py in game app:

    # -*- coding: utf-8 -*-
    from jalka.game.models import Game
    from django.contrib import admin
    
    class GameAdmin(admin.ModelAdmin):
          list_display    = [&#39;type&#39;, &#39;teamone&#39;, &#39;teamtwo&#39;, &#39;gametime&#39;]
          
    admin.site.register(Game, GameAdmin)

project settings.py:

    MIDDLEWARE_CLASSES = (
        &#39;django.middleware.common.CommonMiddleware&#39;,
        &#39;django.contrib.sessions.middleware.SessionMiddleware&#39;,
        &#39;django.middleware.csrf.CsrfViewMiddleware&#39;,
        &#39;django.contrib.auth.middleware.AuthenticationMiddleware&#39;,
        &#39;django.contrib.messages.middleware.MessageMiddleware&#39;,
    )
    
    ROOT_URLCONF = &#39;jalka.urls&#39;
    
    TEMPLATE_DIRS = (
          &quot;/home/projects/jalka/templates/&quot;
    )
    
    INSTALLED_APPS = (
        &#39;django.contrib.auth&#39;,
        &#39;django.contrib.contenttypes&#39;,
        &#39;django.contrib.sessions&#39;,
        &#39;django.contrib.sites&#39;,
        &#39;django.contrib.messages&#39;,
        &#39;django.contrib.admin&#39;,
        &#39;game&#39;,
    )

urls.py:

    from django.conf.urls.defaults import *
    
    # Uncomment the next two lines to enable the admin:
    from django.contrib import admin
    admin.autodiscover()
    
    urlpatterns = patterns(&#39;&#39;,
          # Example:
          # (r&#39;^jalka/&#39;, include(&#39;jalka.foo.urls&#39;)),
          (r&#39;^admin/&#39;, include(admin.site.urls)),
    )

model not showing up in django admin

How can I drop all tables from a database using manage.py and command line? Is there any way to do that executing manage.py with appropriate parameters so I can execute it from a .NET application?

How to drop all tables from the database with manage.py CLI in Django?

Suppose I have a Book model containing a foreign key to a Publisher model.

How can I display in the Django admin a column with the number of books published by each publisher, in a way that I can use the built-in sorting?

 

How to add a sortable count column to the Django admin of a model with a many-to-one relation?

What is the `related_name` argument useful for on `ManyToManyField` and `ForeignKey` fields? For example, given the following code, what is the effect of `related_name=&#39;maps&#39;`?


    class Map(db.Model):
        members = models.ManyToManyField(User, related_name=&#39;maps&#39;,
                                         verbose_name=_(&#39;members&#39;))

What is related_name used for?

I am trying to print a list of all the Conferences and for each conference, print its 3 Speakers.

In my template I have:

    {% if conferences %}
            &lt;ul&gt;
            {% for conference in conferences %}
                    &lt;li&gt;{{ conference.date }}&lt;/li&gt;
                    {% for speakers in conference.speakers %}
                            &lt;li&gt;{{ conference.speakers }}&lt;/li&gt;
                    {% endfor %}
            {% endfor %}
            &lt;/ul&gt;
    {% else %}
    &lt;p&gt;No Conferences&lt;/p&gt;
    {% endif %}

in my views.py file I have:

    from django.shortcuts import render_to_response
    from youthconf.conference.models import Conference
    
    def manageconf(request):
            conferences = Conference.objects.all().order_by(&#39;-date&#39;)[:5]
            return render_to_response(&#39;conference/manageconf.html&#39;, {&#39;conferences&#39;: conferences})

there is a model named **conference**. which has a class named **Conferences** with a *ManyToManyField* named **speakers**

I get the error:

    Caught an exception while rendering: &#39;ManyRelatedManager&#39; object is not iterable

with this line: `{% for speakers in conference.speakers %}`

listing objects from ManyToManyField

There is a handy dynamic attribute in active-record called find_or_create_by:

`Model.find_or_create_by_&lt;attribute&gt;(:&lt;attribute&gt; =&gt; &quot;&quot;)`

But what if I need to find_or_create by more than one attribute?

Say I have a model to handle a M:M relationship between Group and Member called GroupMember. I could have many instances where member_id = 4, but I don&#39;t ever want more than once instance where member_id = 4 and group_id = 7. I&#39;m trying to figure out if it&#39;s possible to do something like this:

    GroupMember.find_or_create(:member_id =&gt; 4, :group_id =&gt; 7)

I realize there may be better ways to handle this, but I like the convenience of the idea of find_or_create.

Rails find_or_create_by more than one attribute?

Ok so this is probably a trivial question but I&#39;m having trouble visualizing and understanding the differences and when to use each. I&#39;m also a little unclear as to how concepts like uni-directional and bi-directional mappings affect the one-to-many/many-to-many relationships. I&#39;m using Hibernate right now so any explanation that&#39;s ORM related will be helpful. 

As an example let&#39;s say I have the following set-up:

    public class Person{
        private Long personId;
        private Set&lt;Skill&gt; skills;
        //Getters and setters
    }
    
    public class Skill{
        private Long skillId;
        private String skillName;
        //Getters and setters
    }

So in this case what kind of mapping would I have? Answers to this specific example are definitely appreciated but I would also really like an overview of when to use either one-to-many and many-to-many and when to use a join table versus a join column and unidirectional versus bidirectional.

Difference Between One-to-Many, Many-to-One and Many-to-Many?

I&#39;m using EF4 and new to it. I have a many to many in my project and cannot seem to work out how to insert or update. I have build a small project just to see how it should be coded.

Suppose I have 3 tables

 1. Class : ClassID-ClassName 
 2. Student : StudentID-FirstName-Surname
 3. StudentClass : StudentID-ClassID

After adding all the relationship and updated the model via model browser I have noticed that StudentClass does not appear, this seems to be default behaviour.

Now I need to do both an Insert and Update. How do you do it? Any code samples or link where I can download an example, or can you spare 5 mins?

Content Type	Original Author	Original Content on Stackoverflow
Question	DZ.	View Question on Stackoverflow
Solution 1 - Django	Ludwik Trammer	View Answer on Stackoverflow

How do I make many-to-many field optional in Django?

Django Problem Overview

Django Solutions

Solution 1 - Django

How to change theme for AlertDialog

Setting Curl's Timeout in PHP

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