按项目所属的类别列出项目

Question

I'm a complete Newb so please bear with me. I've never used OOP or a framework before.

I'm creating a book keeping system where I want the ledgers to be displayed by the category they belong to. Here's my model and it seems to work:

class COAGroup(models.Model):
    Name = models.CharField(max_length=200)
    def __str__(self):
        return self.Name

class Ledger(models.Model):
    Name = models.CharField(max_length=200)
    COAGroup = models.ForeignKey(COAGroup, on_delete=models.CASCADE)
    def __str__(self):
        return self.Name

I'm working on my first view which should list all the ledgers by the category they belong to. Eg

<h2>Some COAGroup Name</h2>
<ul>
 <li>A ledger that belongs to this group</li>
 <li>Another ledger that belonds to the group</li>
 <li>And another</li>
</ul>

<h2>Another COAGroup</h2>
<ul>
 <li>A ledger that belongs to this second group</li>
 <li>Another ledger</li>
</ul>

I've written the following view in views.py:

def showledgers(request):  
    COAGroups = COAGroup.objects.all()
    Ledgers = Ledger.objects.all()
    context = {
        "COAGroups":COAGroups,
        "Ledgers":Ledgers,
    }
    return render(request,"ListLedgers.html",context)

I've managed to get the ledgers to show in ListLedgers.html but I can't figure out how to get them to list by COA Group as per my example.

Thank you!

Answer 1

In your view, you better already .prefetch_related(..) [Django-doc] the related Ledger s:

def showledgers(request):  
    COAGroups = COAGroup.objects.
    context = {
        'groups': COAGroups,
    }
    return render(request, 'ListLedgers.html', context)

Then in the template, you can iterate over the COAGroups , and for each such group over the related .ledger_set :

{% for group in groups %}
  <h2>{{ group.Name }}</h2>
  <ul>
  {% for ledger in  %}
    <li>{{ ledger.name }}</li>
  {% endfor %}
</ul>
{% endfor %}

Note : normally the name of the fields in a Django model are written in snake_case , not PerlCase , so it should be: name instead of Name .