我正在尝试创建一个带有表单的应用程序,该应用程序在提交时会根据提交的信息更新数据库中的表,但我不确定如何进行。目前,我有一个简单的模式:
class Client(models.Model):
company_name = models.CharField(max_length=200)
launchpad_id = models.PositiveIntegerField()
client_email = models.EmailField()
content_id = models.CharField(max_length=200)
def __str__(self):
return self.company_name + ' | ' + self.content_id
我的数据库配置如下:
DATABASES = {
'default': {
'ENGINE': 'django.db.backends.mysql',
'NAME': 'django_project',
'USER': 'xxx',
'PASSWORD': 'xxx',
'HOST': 'xxxx',
'PORT': 'xxx',
},
'info': {
'ENGINE': 'django.db.backends.mysql',
'NAME': 'reporting_database',
'USER': 'xxx',
'PASSWORD': 'xxx',
'HOST': 'xxx',
'PORT': 'xxx',
}
}
我希望发生的情况是,当我通过管理或模板中的客户端模型提交字段时,它会更新reporting_database中的client_info表。不过,我似乎想不出如何让这种联系发挥作用。如果你能给我指引方向,我将不胜感激。谢谢。
发布于 2018-12-06 04:52:36
通过将以下代码添加到我的模型中,我能够获得我想要的结果:
def update_mysql(self):
cursor = db.cursor()
sql = "UPDATE tb_reporting_info SET client_email = '%s' WHERE content_id = '%s' AND launchpad_id = '%s';"
cursor.execute( sql % (self.client_email, self.content_id, self.launchpad_id))
db.commit()
我将表单操作设置为action="{% url 'contact:addClient' %}"
,视图设置为:
def addClient(request):
if request.method == 'POST':
# Set POST data to variables
company_name = request.POST['company_name']
launchpad_id = request.POST['launchpad_id']
content_id = request.POST['content_id']
client_email = request.POST['client_email']
client_added = True # Tells the template to render a success message
# Pass POST data to Client object
c = Client(company_name = company_name,
launchpad_id = launchpad_id,
content_id = content_id,
client_email = client_email
)
c.update_mysql()
它是最基本的,但它完美地满足了我的需求。
发布于 2018-12-01 02:22:11
您需要一个能够创建或更新Model的funx。你可以参考django文档:https://docs.djangoproject.com/en/2.1/ref/models/instances/
发布于 2018-12-01 02:44:08
使用模型表单:
from your_app import Client
class ClientUpdateForm(forms.ModelForm):
class Meta:
model = Client
fields = ('company_name', 'launchpad_id', 'client_email', 'content_id')
def __init__(self, *args, **kwargs):
super(ClientUpdateForm, self).__init__(*args, **kwargs)
self.fields['company_name'].required = True
self.fields['launchpad_id'].required = True
self.fields['client_email'].required = True
self.fields['content_id'].required = True
def clean(self):
cleaned_data = super(ClientUpdateForm, self).clean()
company_name = cleaned_data.get('company_name')
launchpad_id = cleaned_data.get('launchpad_id')
client_email = cleaned_data.get('client_email')
content_id = cleaned_data.get('content_id')
然后从UpdateView继承:
from django.views.generic import UpdateView
from your_app import Client
class ClientUpdateForm(UpdateView):
model = Client
form_class = ClientUpdateForm
template_name_suffix = '_update_form'
def form_valid(self, form):
if form.is_valid():
client = form.save(commit=False)
client.save()
return HttpResponseRedirect('/redirect/')
template_name_suffix意味着您应该在呈现表单client_update_form.html的地方调用模板。
https://stackoverflow.com/questions/53562750
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