簡體   English   中英

Django使用URL參數在基於類的視圖中查詢數據庫

[英]Django using URL parameters to query database in class-based view

我有一個表單,提交后會轉到結果頁面,其中的URL例如:

appname/results/?make=BMW

我想使用URL參數查詢數據庫並在該頁面上顯示結果。

這是我到目前為止的代碼:

意見

class SearchView(FormView):

    template_name = 'carproject/search.html'

    def get(self, request, *args, **kwargs):
        form = AdvancedSearch(self.request.GET or None)
        context = {'form': form}
        if form.is_valid():
            return render(request, 'carproject/results.html')
        else:
            return render(request, self.template_name, context)

class ResultsView(TemplateView):
    template_name = 'carproject/results.html'

    def get_queryset(self, request, *args, **kwargs):
        make = self.request.GET.get('make')
        results = Vehicles.objects.filter(makename__icontains='make')
        context = {'results': results}

        return render(request, self.template_name, context)

HTML

<table>
{% for item in results%}
                <tr>
                    <td>{{item.makename}}</td>
                    <td>{{item.model}}</td>
                    <td>{{item.seriesname}}</td>
                    <td>{{item.seatingcapacity}}</td>
                    <td>{{item.pricenew}}</td>
                </tr>
                {% endfor %}
</table>

我不知道問題是什么,但是你應該改變

results = Vehicles.objects.filter(makename__icontains='make')

results = Vehicles.objects.filter(makename__icontains=make)

'make' -> make

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM