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我只想从模板中调用视图函数,并且我已经看到我必须使用反向函数来完成它。我想将参数传递给视图函数,但我不想在 url 中显示它。可能吗?这是我的代码和错误:

网址.py

url(r'^perfil/deleteDownloadedTrack/$','principal.views.delete_downloaded_track', name='delete_downloaded_track'),

视图.py

def delete_downloaded_track(request, track):    
    ...
    return HttpResponseRedirect(reverse('profile_detail'))

模板.html

{% for track in downloadedTracks %}
    <tr> 
        <td>
            {{track.name}} 
        </td>
        <td> 
            <div class="botoForm" onclick="location.href='{% url "principal.views.delete_downloaded_track" track %}'"> Eliminar </div>
        </td>
{% endfor %}

和错误:

Reverse for 'principal.views.delete_downloaded_track' with arguments '(<DownloadedTrack: DownloadedTrack object>,)' and keyword arguments '{}' not found.

谢谢!!

4

2 回答 2

2

HTML 标准推荐的处理任何更改服务器数据的请求的方法是使用 POST 请求。

{% for track in downloadedTracks %}
    <tr> 
        <td>
            {{track.name}} 
        </td>
        <td> 
            <form method="POST" action="{% url principal.views.delete_downloaded_track %}" id="delete-form-{{ track.id }}">
                {% csrf_token %}
                <input type="hidden" name="track_id" value="{{ track.id }}" />
                <div class="botoForm" onclick="document.getElementById('delete-form-{{ track.id }}').submit()"> Eliminar </div>
            </form>
        </td>
{% endfor %}

通过验证您将此表单发送给的人也是通过 POST 请求将其发回的人,{% csrf_token %}从而防止跨站点伪造请求。只需使用以下命令检索视图中的 id:

if request.method == 'POST':
    id = request.POST.get("track_id", None)
    track = Track.objects.get(id=id)

通过 POST 发送的数据不会显示在 url 中。

于 2013-07-27T18:34:22.863 回答
1

如果您不想在 URL 中显示参数,那么您可能希望将其作为查询参数传递,例如

<host_and_path>/perfil/deleteDownloadedTrack/?track_id=<some_track_id>.

因此,您的视图将如下所示:

def delete_downloaded_track(request):
    track_id = request.GET.get('track_id')
    # get track based on track_id and delete it.

和你的模板是这样的(假设你在模板上下文中将 delete_downloaded_track 作为 delete_url 传递):

{% for track in downloadedTracks %}

    <!-- template code -->

    {% with delete_url|add:"?track_id="|add:track.id as url_with_id %}
        <a href="{{url_with_id|safe}}">Delete</a>
    {% endwith %}

{% endfor %}
于 2013-07-27T17:32:22.390 回答