Question

现在的问题是，通过Jinja，我想在html中调情数据，有两家公司，我希望一家公司在另一个地方展示，而第二家在另一个地方展示

我被困在这里，这里显示了所有内容，但是我想按公司进行过滤，我知道views.py可以完成，但是我想通过jinja寻求帮助

models.py

from django.db import models

class Company(models.Model):
    name = models.CharField(max_length=200)

    def __str__(self):
        return self.name

class Buses(models.Model):
    company = models.ForeignKey(Company, on_delete=models.DO_NOTHING)
    name = models.CharField(max_length=200)
    state_number = models.CharField(max_length=200)
    route = models.IntegerField()
    phone = models.IntegerField(blank=True)
    paid_until = models.IntegerField(blank=True)
    photo_person = models.ImageField(upload_to='photos/%Y/%m/%d/', default='avatar.jpg')
    photo_front = models.ImageField(upload_to='photos/%Y/%m/%d/', default='carcas.png')
    photo_back = models.ImageField(upload_to='photos/%Y/%m/%d/', default='carcas.png')
    photo_car_pass = models.ImageField(upload_to='photos/%Y/%m/%d/', default='carcas.png')
    photo_license = models.ImageField(upload_to='photos/%Y/%m/%d/', default='carcas.png')
    photo_route = models.ImageField(upload_to='photos/%Y/%m/%d/', default='carcas.png')

    def __str__(self):
        return self.name

index.html

<ul class="list-group collapse show" id="collapseM">
    {% for bus in buses %}
    <li class="list-group-item">
        <div class="row">
            <div class="col-lg-4">
                <a class="" href='#'><b>{{bus.name}}</b></a>
            </div>
            <div class="col-lg-4">
                {{bus.state_number}}
            </div>
            <div class="col-lg-4">
                {{bus.route}}
            </div>
        </div>
    </li>
    {% endfor %}
</ul>

views.py

from django.shortcuts import get_object_or_404, render
from .models import Buses

def index(request):
    buses = Buses.objects.all()

    context = {
        'buses': buses
    }

    return render(request, 'buses/index.html', context)

Jinja / Django如何从Jinja过滤？

0 个答案: