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Django按第一个字母分组查询集?

django-templates sorting django

Tudor · 技术社区 · 15 年前

我有一个查询集,比如:

items = Item.objects.all()

项有一个“名称”字段。在我要显示的模板中:

一
轴线
酒
乙
火箭筒
C
硬币
弹药筒
S
剑
麻雀

所以项目是按第一个字母排序和分组的。漏字母省略。有人有什么想法吗?

6 回复 | 直到 7 年前

Elf Sternberg 15 年前

class Item(models.Model):
    ...

    def first_letter(self):
        return self.name and self.name[0] or ''

{% 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>

Ross 13 年前

return self.name and self.name.upper()[0] or ''

Ollie Cook 11 年前

slice first_letter

{% 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>

user2633396 9 年前

{% 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 item.name[0]

Basil Jose 7 年前

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)

class City(models.Model):
    """All features model"""

    name = models.CharField(max_length=99)

{
    "A": [
        "Adelanto",
        "Azusa",
        "Alameda",
        "Albany",
        "Alhambra",
        "Anaheim"
    ],
    "B": [
        "Belmont",
        "Berkeley",
        "Beverly Hills",
        "Big Sur",
        "Burbank"
    ],
    ......

}

Javier Buzzi 7 年前

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

collector yield