在 Django 中的 AJAX POST 请求后重新加载模板

Question

我正在使用 HTML5 地理定位来获取用户的位置，然后将纬度/经度发送到我的 django 应用程序以查找最近的三所学校。我可以发布纬度/经度，通过函数运行它以获取最近的学校，并在终端中打印出 dict 对象，但模板永远不会重新加载新的 context_dict 数据。

HTML

{% csrf_token %}
<input id = 'button' type="submit" value="Use current location" onclick = 'find_school()' class="btn btn-default">

JS

function find_school(){

function send_off(lat, long){
    var locale = [lat, long];
    console.log(lat, long);
    return locale
}

navigator.geolocation.getCurrentPosition(function(position) {
    $.ajax({
        type: "POST",
        url: "/schools/search/",
        data: {
            csrfmiddlewaretoken: document.getElementsByName('csrfmiddlewaretoken')[0].value,
            lat_pos: position.coords.latitude,
            long_pos: position.coords.longitude
        },
        success: function(data){
            console.log(data);
        }
    })
});

}

views.py

def find_school(request):
    context = RequestContext(request)
    search_school_list = search_school_bar()
    if request.GET:
        address = request.GET['q_word']
        close_schools = geolocate(address)
        context_dict = {'close_schools': close_schools, 'search_schools':json.dumps(search_school_list)}
    elif request.method == 'POST' and request.is_ajax():
        position = request.POST

        #the geo_search view takes the lat and long
        return geo_search(request, position['lat_pos'],position['long_pos'] )
    else:
        context_dict = {'search_schools':json.dumps(search_school_list)}

    return render_to_response('school_data/search.html', context_dict, context)

def geo_search(request, lat, long):
    context = RequestContext(request)
    search_school_list = search_school_bar()

    close_schools = geolocate_gps(lat, long)

    context_dict = {'close_schools': close_schools, 'search_schools':json.dumps(search_school_list)}
    #This print statement returns in my terminal the results, and they are correct. I just need to reload the template to display the results.
    print context_dict
    #This isn't re-rendering the page with the correct context_dict. It is doing nothing.
    return render_to_response('school_data/search.html', context_dict, context)

Answer 1

如果您将 render_to_response 的输出返回给 ajax 调用，您将在 javascript 调用中将 HTML 作为“success:function(data)”中的“data”元素返回。如果您的目标是重新加载页面，我认为您不想使用 AJAX。

你应该有类似的东西：

navigator.geolocation.getCurrentPosition(function(position) {
 var form = $('<form action="/geosearch/" method="POST"></form>');
 var long =$('<input name = "long" type="hidden"></input>');
 var lat =$('<input name = "lat" type="hidden"></input>');
 lat.val(position.coords.latitude);
 long.val(position.coords.latitude);
 form.append(lat, long);
 $('body').append(form);
 form.submit();
}

您在 /geosearch/ 的视图应该采用 post 变量，做任何您想做的事情，然后 render_to_response。