22👍
There’s a template tag for this, if all you care about is its presentation on the page. First, define an organizational principle in the class. In your case, it’s the first letter:
class Item(models.Model):
...
def first_letter(self):
return self.name and self.name[0] or ''
And then define a regroup in the template, using the first_letter call:
{% regroup items by first_letter as letter_list %}
<ul>
{% for letter in letter_list %}
<li>{{ letter.grouper }}
<ul>
{% for item in letter.list %}
<li>{{ item.name }}</li>
{% endfor %}
</ul>
</li>
{% endfor %}
</ul>
7👍
Just wanted to add that if you use this and your item has a lower-case first character it will be a separate group. I added upper to it.
return self.name and self.name.upper()[0] or ''
5👍
Alternatively you could use slice inline in the template without the need for a first_letter method on your model.
{% regroup items by name|slice:":1" as letter_list %}
<ul>
{% for letter in letter_list %}
<li>{{ letter.grouper }}
<ul>
{% for item in letter.list %}
<li>{{ item.name }}</li>
{% endfor %}
</ul>
</li>
{% endfor %}
</ul>
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1👍
Even easier. You can group by first leter just in ‘regroup’:
{% regroup items|dictsort:"name" by name.0 as item_letter %}
<ul>
{% for letter in item_letter %}
<h4>{{ letter.grouper|title }}</h4>
{% for i in letter.list|dictsort:"name" %}
<li>{{ i.name }}</li>
{% endfor %}
{% empty %}
<span>There is no items yet...</span>
{% endfor %}
</ul>
name.0 in this case the same as item.name[0] in Python.
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1👍
For Django REST you can do like this,
import string
import collections
from rest_framework.response import Response
from rest_framework import status, viewsets
def groupby(self, request):
result = []
for i in list(string.ascii_uppercase):
c = City.objects.filter(name__startswith=i)
if c:
result.append((i, map((lambda x: x['name']),list(c.values('name')))
))
return Response(collections.OrderedDict(sorted(dict(result).items())), status=status.HTTP_200_OK)
City Models
class City(models.Model):
"""All features model"""
name = models.CharField(max_length=99)
Response
{
"A": [
"Adelanto",
"Azusa",
"Alameda",
"Albany",
"Alhambra",
"Anaheim"
],
"B": [
"Belmont",
"Berkeley",
"Beverly Hills",
"Big Sur",
"Burbank"
],
......
}
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0👍
This is another take for doing in straight Django and Python. The other solution offered was terribly inefficient
import itertools
collector = {}
item_qs = Item.objects.all().order_by('name')
for alphabet_letter, items in itertools.groupby(item_qs, lambda x: x.name[0].lower()):
# you can do any modifications you need at this point
# you can do another loop if you want or a dictionary comprehension
collector[alphabet_letter] = items
What does this give you? A single query to the db.
Should I use a collector? No, you should maybe use a yield this is just a proof of concept.
What ever you do, DO NOT add a query call inside a loop.