Thankyou, Daniel - that's done the trick. I can't believe I didn't
think of that before.
Thanks for the point about aggregation, too - I suspected the way I
was doing things was probably not the most efficient, and I'll
certainly look into the alternative.
Bianca
On Jul 29, 5:17 pm, Daniel Roseman <dan...@roseman.org.uk> wrote:
> On Jul 29, 5:18 am, bweiss <weissmanbia...@gmail.com> wrote:
>
>
>
> > I've got this to work in python shell mode, but am having trouble
> > translating it into a django view, and am hoping someone can point out
> > where I'm stuffing up.
>
> > For background, I'm working with a database with two models:
> > "Employees" and "Projects", joined by a ForeignKey field on the
> > Project model called "EmployeeName". Each project has exactly one
> > employee associated with it, and also has a field called "Type" with
> > options of 1, 2, or 3.
>
> > I need to write a view that will generate a list of employees and
> > then, for each employee in that list, generate a related list of all
> > Type 1 projects under their name, a list of Type 2 projects under
> > their name, and a list of Type 3 projects under their name.
>
> > (The reason I have to do it this way, rather than simply generating
> > one big list of all projects and then just re-ordering by name and
> > type, is that I want to be able to cycle through each list separately
> > in order to do some template-side manipulation of the way the data is
> > presented. Eg. control how many projects of each type per person are
> > presented, what to display if an employee has no projects of a certain
> > type, etc etc. I know how to do this with the template tags, but am
> > having trouble getting the template to access the data lists.)
>
> > I've got the query to work in shell mode like so:
>
> > >> for employee in Employee.objects.filter([filters here]):
> > >> a =
> > >> Project.objects.filter(Type=1).filter(EmployeeName__id=employee.id)
> > >> b =
> > >> Project.objects.filter(Type=2).filter(EmployeeName__id=employee.id)
> > >> c =
> > >> Project.objects.filter(Type=3).filter(EmployeeName__id=employee.id)
> > >> print e.Name, a, b, c
>
> > That works fine and displays exactly the dataset I want. However, I'm
> > having trouble translating that into the django view.
>
> > At the moment, my view looks like this:
>
> > def employeeprojectreport(request):
> > employee_list = Employee.objects.filter([filters here])
> > for employee in employee_list:
> > a_list = Project.objects.filter(Type=1).filter
> > (EmployeeName__id=employee.id)
> > b_list = Project.objects.filter(Type=2).filter
> > (EmployeeName__id=employee.id)
> > c_list = Project.objects.filter(Type=3).filter
> > (EmployeeName__id=employee.id)
>
> > t = loader.get_template('employeeprojectreport.html')
> > c = Context({
> > 'employee_list': employee_list,
> > 'a_list': a_list,
> > 'b_list': b_list,
> > 'c_list': c_list,
>
> > })
> > return HttpResponse(t.render(c))
>
> > The template is cycling through employee_list the way I want it to,
> > but I can't seem to call the related lists through the template.
> > Either I'm defining things wrongly in the view, or doing something
> > wrong at the template end.
>
> > In the template, I've tried:
> > {% for employee in employee_list %}
> > {{ a }}
> > {% endfor %}
>
> > and:
> > {% for employee in employee_list %}
> > {% for a in a_list %}
> > {{ a }}
> > {% endfor %}{% endfor %}
>
> > and:
> > {% for employee in employee_list %}
> > {% for a in employee.a_list %}
> > {{ a }}
> > {% endfor %}{% endfor %}
>
> > The first outputs an empty set of square brackets [], the other two
> > don't output anything.
>
> > Does anyone have any suggestions as to where I might be going wrong?
>
> > Many thanks,
>
> > Bianca
>
> Your view isn't defining one of each list per employee - it's defining
> a single a_list, b_list and c_list, and overwriting it each time. So,
> by the time you reach the end of the employee list, your three lists
> just contain the data for the final employee.
>
> One way of fixing this is to assign each list as an attribute of the
> relevant employee:
> for employee in employee_list:
> employee.a_list = Project.objects.filter(Type=1).filter(
> EmployeeName__id=employee.id)
> etc.
> Then in your templates you can do {% for a in employee.a_list %} etc.
>
> However, I have to say that this is an extremely inefficient method of
> doing the lookups - you're generating three database calls per
> employee. If you have hundreds of employees you'll soon run into
> performance issues. You may want to investigate the aggregation
> features in version 1.1 - that will allow you to group your projects
> by project type, which will probably be more efficient.
> --
> DR.
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