Django中AJAX POST请求后重新加载模板

Reload template after AJAX POST request in Django

本文关键字：加载 新加载 AJAX POST 请求 Django      更新时间：2023-09-26

我使用HTML5地理定位来获取用户的位置，然后将lat/long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lat/long，通过一个函数运行它来获得最近的学校，并在终端中打印出dict对象，但模板从未重新加载新的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)

如果将render_to_response的输出返回到ajax调用，则在javascript调用中返回HTML作为"success:function（data）"中的"data"元素。如果你的目标是重新加载页面，我认为你不想使用AJAX。

你应该有这样的东西：

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();
}

你在/geosearch/上的视图应该采用post变量，做你想做的任何事情，然后render_to_response。

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