2016-07-22 44 views
0

我試圖按照這裏給出的例子:Refresh <div> element generated by a django templateNoReverseMatch {%URL%}

但我結束了:

NoReverseMatch at /search/: 
Reverse for '' with arguments '()' and keyword arguments 
'{u'flights':[{FLIGHTS}]}' not found. 0 pattern(s) tried: []" 

我看到很多搜索結果的這個錯誤,但除非我完全錯過了某些東西,否則它們都不相關。

的Javascript在search.html:(修訂版)

 <script> 
     var flights = {{ flights | safe }} 

     $.ajax({ 
     url: {% url 'search_results' flights %}, 
     success: function(data) { 
     $('#search-results').html(data); 
     } 
     }); 
     </script> 

views.py:

def search_results(request, flights): 
    return render_to_response('search_results.html', flights) 

urls.py:(修訂版)

url(r'^search/search_results/(?P<flights>[^/]*)$', "fsApp.views.search_results", name='search_results'), 

ETA:

我現在嘗試了以下所有內容,並且無工作:

url(r'^search/(?P<flights>[^/]*)$', "fsApp.views.search_results", name='search_results'),              
url(r'^search/search_results/(?P<flights>[^/]*)$', "fsApp.views.search_results", name='search_results'), 
url(r'^search/(?P<flights>[^/]*)/search_results/$', "fsApp.views.search_results", name='search_results'), 
+1

的search_results網址未任何參數,但你試圖通過它他看重「航班」。 –

+0

我已經嘗試了三種不同的方式(基於響應和文檔),並且它們都不起作用。 – ballardjw2

回答

1

你缺少URL的名字:

url(r'^search/(?P<flights>[^/]*)/search_results/$', "fsApp.views.search_results", name='search_results'), 

在模板:

{% url 'search_results' flights as url_search %} 

<script> 
    ... 
    url: '{{ url_search }}', 
    ... 
</script> 

Example from Django

from django.conf.urls import url 

from . import views 

urlpatterns = [ 
    #... 
    url(r'^articles/([0-9]{4})/$', views.year_archive, name='news-year-archive'), 
    #... 
] 
+0

修復了我的代碼,但我得到了同樣的錯誤。 – ballardjw2

+0

您目前的網址定義並不期待'航班'參數 – Gocht

+0

@ ballardjw2我編輯了我的答案,並在您的網址中請求一個kwarg的示例,查看我的答案中鏈接的文檔以瞭解更多信息。 – Gocht