我有这个代码(它没有给我预期的结果)
#subject_content.html
{% block main-menu %}
{% include "subject_base.html" %}
{% endblock %}
#subject_base.html
....
....
<div id="homework" class="tab-section">
<h2>Homework</h2>
{% include "subject_file_upload.html" %}
</div>
子模板:
#subject_file_upload.html
<form action="." method="post" enctype="multipart/form-data">{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="submit">
</form>
我的观点是
#views.py
@login_required
def subject(request,username, subject):
if request.method == "POST":
form = CarsForm(request.POST, request.FILES)
if form.is_valid():
form.save()
return HttpResponseRedirect("/")
form = CarsForm()
return render_to_response('subject_content.html', {'form':form}, context_instance=RequestContext(request))
上面的代码按照我想要的方式创建HTML,但是表单不会更新数据库。
但,
如果我跳过中间的模板,直接转到上传表单,它工作得很好:
#subject_content.html
{% block main-menu %}
{% include "subject_file_upload.html" %}
{% endblock %}
请帮我让它与中间模板一起工作。我想这么做,因为我不想多次输入相同的代码。
发布于 2012-07-25 05:36:14
就像@Besnik建议的那样,这很简单:
{% include "subject_file_upload.html" with form=form foo=bar %}
documentation for include
提到了这一点。它还提到,您可以使用only
来呈现仅具有给定变量的模板,而不继承任何其他变量。
谢谢@Besnik
https://stackoverflow.com/questions/11639306
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