Consultas sql sin procesar en vistas de Django

Question

¿Cómo realizaría lo siguiente con sql sin formato en views.py?

from app.models import Picture 

def results(request): 
    all = Picture.objects.all() 
    yes = Picture.objects.filter(vote='yes').count() 
    return render_to_response('results.html', {'picture':picture, 'all':all, 'yes': yes}, context_instance=RequestContext(request)) 

¿Cómo sería esta función de resultados? Gracias.

Answer 1

>>> from django.db import connection 
>>> cursor = connection.cursor() 
>>> cursor.execute('''SELECT count(*) FROM people_person''') 
1L 
>>> row = cursor.fetchone() 
>>> print row 
(12L,) 
>>> Person.objects.all().count() 
12 

uso cláusula WHERE para filtrar votar por sí:

>>> cursor.execute('''SELECT count(*) FROM people_person WHERE vote = "yes"''') 
1L 

Answer 2

The Django Documentation is really really good. Tiene básicamente dos opciones para ejecutar SQL sin formato. Puede usar Manager.raw() para realizar consultas sin procesar que devuelven instancias del modelo, o puede evitar la capa del modelo y ejecutar SQL personalizado directamente.

Utilizando el raw() gerente:

>>> for p in Person.objects.raw('SELECT * FROM myapp_person'): 
...  print p 
John Smith 
Jane Jones 

Si desea omitir la capa de modelo directamente puede utilizar django.db.connection que representa la conexión de base de datos por defecto:

def my_custom_sql(): 
    from django.db import connection, transaction 
    cursor = connection.cursor() 

    # Data modifying operation - commit required 
    cursor.execute("UPDATE bar SET foo = 1 WHERE baz = %s", [self.baz]) 
    transaction.commit_unless_managed() 

    # Data retrieval operation - no commit required 
    cursor.execute("SELECT foo FROM bar WHERE baz = %s", [self.baz]) 
    row = cursor.fetchone() 

    return row 

Answer 3

puede probar esta

Picture.objects.raw("SELECT 1 as id ,"\ 
"(SELECT count(*) as yes FROM people_person WHERE vote='yes') as yes ,"\ 
"(SELECT count(*) FROM people_person WHERE vote='no') as no ,"\ 
"(SELECT count(*) FROM people_person WHERE vote='all') as all ")