首页 > 解决方案 > 尝试导入 json 时如何修复 django 中的 ImportError?

问题描述

我有一个网页,其中包含 2 个链接的下拉列表国家和城市,其中基于第一个下拉列表的选择,第二个下拉列表显示城市属于所选国家/地区。

问题是第一个下拉列表显示从数据库中获取的数据,但第二个仍然是空的。并且系统显示以下错误:

from django.utils import json as simplejson ImportError: cannot import name 'json' from 'django.utils'

模型.py

from django.db import models
    class country(models.Model):
        name = models.CharField(max_length=100)

        def __str__(self):
            return str(self.name)

    class city(models.Model):
        name = models.CharField(max_length=100)
        MouhafazatID = models.ForeignKey(country,on_delete=models.CASCADE)


        def __str__(self):
            return str(self.name)

网址.py

from django.contrib import admin
from django.urls import path, include
from.views import *

urlpatterns = [
    path('admin/', admin.site.urls),
    # path('', home),
    path('', home2),
    path('getdetails/', getdetails),

视图.py

from django.shortcuts import render
from django.http import HttpResponse
from testapp.models import *

from django.utils import json as simplejson # i think this is the error?



def home2(request):
    countries = country.objects.all()
    print(countries)
    return render(request, 'home2.html',{'countries': countries})



def getdetails(request):

    #country_name = request.POST['country_name']
    country_name = request.GET['cnt']
    print ("ajax country_name ", country_name)

    result_set = []
    all_cities = []

    answer = str(country_name[1:-1])
    selected_country = country.objects.get(name=answer)
    print ("selected country name ", selected_country)

    all_cities = selected_country.city_set.all()
    for city in all_cities:
        print ("city name", city.name)
        result_set.append({'name': city.name})

    return HttpResponse(simplejson.dumps(result_set), mimetype='application/json', content_type='application/json')

最后一行出现错误如何解决?

home2.html

<html>
    <head>
    <script type="text/javascript" src="http://code.jquery.com/jquery-latest.min.js"></script>
    <script type="text/javascript" src="http://yourjavascript.com/7174319415/script.js"></script>
        <script>
            $(document).ready(function(){
                 $('select#selectcountries').change(function () {
                     var optionSelected = $(this).find("option:selected");
                     var valueSelected  = optionSelected.val();
                     var country_name   = optionSelected.text();


                     data = {'cnt' : country_name };
                     ajax('/getdetails',data,function(result){

                            console.log(result);
                            $("#selectcities option").remove();
                            for (var i = result.length - 1; i >= 0; i--) {
                                $("#selectcities").append('<option>'+ result[i].name +'</option>');
                            };


                         });
                 });
            });
        </script>
    </head>

    <body>
        <select name="selectcountries" id="selectcountries">
        {% for item in countries %}
            <option val="{{ item.name }}"> {{ item.name }} </option>    
        {% endfor %}
        </select>   


        <select name ="selectcities" id="selectcities">


        </select>

    </body>
</html>

标签: ajaxdjangosqlitedropdown

解决方案


我能够通过更改

from django.utils import json as simplejson

import json as simplejson

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