我无法让提交与crispy-froms一起使用。正常的django形式与bootstrap工作正常。我已经尝试了所有可以找到的教程,但此时无法找到代码有什么问题。
点击sumbit时会打开我的客户概述页面,但没有添加新客户。此处并未显示所有字段,但字段设置均设置为允许空值。
我的models.py
from django.db import models
from django.utils.encoding import smart_unicode
class CustomerType(models.Model):
customer_type = models.CharField(max_length=120, null=True, blank=True)
timestamp_created = models.DateTimeField(auto_now_add=True, auto_now=False)
timestamp_updated = models.DateTimeField(auto_now_add=False, auto_now=True)
def __unicode__(self):
return smart_unicode(self.customer_type)
class Customer(models.Model):
customer_type = models.ForeignKey(CustomerType, null=True, blank=True)
customer_name = models.CharField(max_length=120, null=True, blank=True)
my views.py
def customercrispy(request):
form = ExampleForm()
return render_to_response("customer-crispy.html",
{"example_form": form},
context_instance=RequestContext(request))
my forms.py
from django import forms
from crispy_forms.helper import FormHelper
from crispy_forms.layout import Layout, Fieldset, ButtonHolder, Submit, Div, Field
from crispy_forms.bootstrap import TabHolder, Tab, FormActions
from .models import Customer
class CustomerAddForm(forms.ModelForm):
class Meta:
model = Customer
class ExampleForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
super(ExampleForm, self).__init__(*args, **kwargs)
self.helper = FormHelper(self)
self.helper.form_method = 'post'
self.helper.form_action = '/customeroverview/'
self.helper.add_input(Submit('submit', 'Submit'))
class Meta:
model = Customer
编辑:打开表单时的CMD输出
[08/Oct/2014 12:45:12] "GET /customercrispy/ HTTP/1.1" 200 11203
[08/Oct/2014 12:45:12] "GET /customercrispy/static/js/ie-emulation-modes-warning.js HTTP/1.1" 404 3118
[08/Oct/2014 12:45:12] "GET /assets/js/ie10-viewport-bug-workaround.js HTTP/1.1" 404 3079
保存时的CMD输出
[08/Oct/2014 12:46:52] "POST /customeroverview/ HTTP/1.1" 200 5129
[08/Oct/2014 12:46:52] "GET /customeroverview/static/js/ie-emulation-modes-warning.js HTTP/1.1" 404 3124
[08/Oct/2014 12:46:52] "GET /assets/js/ie10-viewport-bug-workaround.js HTTP/1.1" 404 3079
答案 0 :(得分:1)
您的观点不合适。目前,您只需创建一个表单并进行渲染。发布表单时没有任何操作,只是再次呈现。因此,如果是POST,请检查表单是否有效并保存。所以这将使它像:
def customercrispy(request):
if request.method == 'POST':
form = ExampleForm(request.POST)
if form.is_valid():
form.save()
return redirect('customercripsy') # name of view stated in urls
else:
form = ExampleForm()
return render_to_response("customer-crispy.html",
{"example_form": form},
context_instance=RequestContext(request))
这也可以避免您设置' / customeroverview /'作为形式行动。如果您确实要发布到此网址,请在您的问题中添加customeroverview
视图。顺便说一句,我建议你使用django的reverse
函数来创建你的网址。 (https://docs.djangoproject.com/en/dev/ref/urlresolvers/#reverse)。
有关视图中的模型形式的更多文档:https://docs.djangoproject.com/en/dev/topics/forms/modelforms/#using-a-model-formset-in-a-view