Django - 使用 ManyToMany 保存表单集答案

【问题标题】：Django - saving formset with ManyToManyDjango - 使用 ManyToMany 保存表单集
【发布时间】：2012-09-27 23:02:25
【问题描述】：

我遇到了以下问题。一旦我尝试保存已编辑的对象（详情如下），就会出现错误：

MultiValueDictKeyError at /apps/edit/1/

"Key 'application2server_set-0-id' not found in <QueryDict: {u'application2server_set-MAX_NUM_FORMS': [u''], u'name': [u'application1'], u'repository': [u'1'], u'application2server_set-INITIAL_FORMS': [u'2'], u'application2server_set-TOTAL_FORMS': [u'5'], u'csrfmiddlewaretoken': [u'bmEPdLCloNHKR0qUmwdhdmQ4aDKHV2CT'], u'servers': [u'1', u'2']}>"

以下是使用的模型和形式：

class Server(models.Model):
    host = models.CharField(max_length=100)
    user = models.CharField(max_length=20)
    root_directory = models.CharField(max_length=200)


class Repository(models.Model):
    host = models.CharField(max_length=100)
    user = models.CharField(max_length=20)


class Application(models.Model):
    name = models.CharField(max_length=100)
    repository = models.ForeignKey(Repository)
    servers = models.ManyToManyField(Server, through='Application2Server', null=True, blank=True)


class Application2Server(models.Model):
    application = models.ForeignKey(Application)
    server = models.ForeignKey(Server)
    tag = models.CharField(max_length=100)
    deployment_date = models.DateTimeField(auto_now=True, blank=True)


class ServerForm(ModelForm):
    class Meta:
        model = Server


class RepositoryForm(ModelForm):
    class Meta:
        model = Repository


class ApplicationForm(ModelForm):
    class Meta:
        model = Application


class Application2ServerForm(ModelForm):
    class Meta:
        model = Application2Server
         exclude = ('application', 'server', 'tag')

查看用于保存日期：

def app_edit(request, id):
    application = get_object_or_404(Application, pk=id)
    Application2ServerFormSet = inlineformset_factory(Application, Application2Server)

    if request.method == 'POST':
        form = ApplicationForm(request.POST, instance=application)
        formset = Application2ServerFormSet(request.POST, instance=application)

        if form.is_valid() and formset.is_valid():
            saved_application = form.save()
            formset.save()

            return HttpResponseRedirect(reverse(
                'proj.views.app_show',
                args=(saved_application.pk,)
            ))
    else:
        form = ApplicationForm(instance=application)
        formset = Application2ServerFormSet(instance=application)

    return render(request, 'app_edit.html', {'form': form, 'formset': formset})

和模板来生成 HTML 和表单本身：

<h1>App edit view</h1>
{% extends "base.html" %}

{% block container %}
    <form method="post">{% csrf_token %}
        <table>
            {{ formset.management_form }}

            {{ form.as_table }}
            <tr><th></th><td><input type="submit" value="Save changes" /></td></tr>
        </table>
    </form>
{% endblock %}

任何帮助将不胜感激！谢谢！

【问题讨论】：

标签： python django forms save formset

【解决方案1】：

您需要在表单集中以及在模板中呈现每个表单。

将您的模板更新为这样的内容（您正在渲染表单集的地方）：

<form method="post" action="">
    {{ formset.management_form }}
    <table>
        {% for form in formset %}
        {{ form.as_table }}
        {% endfor %}
    </table>
</form>

参考：Formset in templates

【讨论】：