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我有這樣一個模型:Django的多個查詢過濾器 - 1場或2場
class Result(models.Model):
title = models.CharField(max_length=300)
desc = models.TextField(max_length=800)
url = models.CharField(max_length=200)
rank = models.BigIntegerField()
category = models.ForeignKey('Category')
tags = TaggableManager()
def __str__(self):
return self.title
class Meta:
verbose_name_plural = "Risultati"
class Category(models.Model):
title = models.CharField(max_length=300)
slug = models.SlugField(null=True)
def __str__(self):
return self.title
class Meta:
verbose_name_plural = "Categorie"
和看法:
def search_form(request):
return render(request, 'search.html')
def result(request):
if 'q' in request.GET and request.GET['q']:
q = request.GET['q']
results = Result.objects.filter(tags__name__in=[q], desc__icontains=q).order_by('-rank')
return render(request, 'results.html', {'results': results, 'query': q})
else:
return render(request, 'search.html', {'error': True})
def detail(request, result_id):
try:
result = Result.objects.get(pk=result_id)
except Result.DoesNotExist:
raise Http404("Nessun risultato :(")
return render(request, 'detail.html', {'result': result})
我需要通過過濾器,標籤或遞減搜索,我嘗試連接過濾器()。filter()但沒有發生,錯誤在哪裏?
謝謝你,完美! – user0111001101