我正在使用django 1.10
创建我的个人网站以下是技能应用的模型:
from __future__ import unicode_literals
from django.db import models
# Create your models here.
class Skill(models.Model):
name = models.CharField(max_length=256)
created_at = models.DateTimeField(auto_now=False, auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True, auto_now_add=False)
def __unicode__(self):
return self.name
def __str__(self):
return self.name
class Subskill(models.Model):
skill = models.ForeignKey(Skill, on_delete=models.CASCADE)
name = models.CharField(max_length=256)
link = models.CharField(max_length=256)
created_at = models.DateTimeField(auto_now=False, auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True, auto_now_add=False)
def __unicode__(self):
return self.name
def __str__(self):
return self.name
并查看:
from django.shortcuts import render
from skill.models import Skill,Subskill
# Create your views here.
def home(request):
skill = Skill.objects.all()
subskill =Subskill.objects.all()
context = {'skills':skill,
'subskills':subskill}
return render(request, 'skill.html', context)
这是我的模板页面:
{% block skill %}
{% for subskill in subskills %}
{{subskill.skill.name}}
{{subskill.name}}
{% endfor %}
{% endblock skill %}
假设有一项名为网页设计的技能,它有两个名为html和css的子项。 我想在视图页面中呈现技能名称和它的两个子名称:
网页设计
HTML
CSS
但它呈现像Web设计Html Web设计CSS
请帮我解决这个问题。
答案 0 :(得分:4)
你可以对技能本身进行realted query https://docs.djangoproject.com/en/1.10/topics/db/queries/#backwards-related-objects
# example
skill_obj = Skill.objects.all()[0]
subskills = skill_obj.subskill_set.all()
或者在你的情况下
def home(request):
skills = Skill.objects.all().prefetch_related('subskill_set') # optimizing
context = {'skills':skills}
return render(request, 'skill.html', context)
在模板中
{% for skill in skills %}
{{skill.name}}
{% for subskill in skill.subskill_set.all %}
{{subskill.name}}
{% endfor %}
{% endfor %}